5840

44
+67  Q:

## Challenge

Here is the challenge (of my own invention, though I wouldn't be surprised if it has previously appeared elsewhere on the web).

Write a function that takes a single argument that is a string representation of a simple mathematical expression and evaluates it as a floating point value. A "simple expression" may include any of the following: positive or negative decimal numbers, +, -, *, /, (, ). Expressions use (normal) infix notation. Operators should be evaluated in the order they appear, i.e. not as in BODMAS, though brackets should be correctly observed, of course. The function should return the correct result for any possible expression of this form. However, the function does not have to handle malformed expressions (i.e. ones with bad syntax).

Examples of expressions:

``````1 + 3 / -8                            = -0.5       (No BODMAS)
2*3*4*5+99                            = 219
4 * (9 - 4) / (2 * 6 - 2) + 8         = 10
1 + ((123 * 3 - 69) / 100)            = 4
2.45/8.5*9.27+(5*0.0023)              = 2.68...
``````

## Rules

I anticipate some form of "cheating"/craftiness here, so please let me forewarn against it! By cheating, I refer to the use of the `eval` or equivalent function in dynamic languages such as JavaScript or PHP, or equally compiling and executing code on the fly. (I think my specification of "no BODMAS" has pretty much guaranteed this however.) Apart from that, there are no restrictions. I anticipate a few Regex solutions here, but it would be nice to see more than just that.

Now, I'm mainly interested in a C#/.NET solution here, but any other language would be perfectly acceptable too (in particular, F# and Python for the functional/mixed approaches). I haven't yet decided whether I'm going to accept the shortest or most ingenious solution (at least for the language) as the answer, but I would welcome any form of solution in any language, except what I've just prohibited above!

## My Solution

I've now posted my C# solution here (403 chars). Update: My new solution has beaten the old one significantly at 294 chars, with the help of a bit of lovely regex! I suspected that this will get easily beaten by some of the languages out there with lighter syntax (particularly the funcional/dynamic ones), and have been proved right, but I'd be curious if someone could beat this in C# still.

## Update

I've seen some very crafty solutions already. Thanks to everyone who has posted one. Although I haven't tested any of them yet, I'm going to trust people and assume they at least work with all of the given examples.

Just for the note, re-entrancy (i.e. thread-safety) is not a requirement for the function, though it is a bonus.

## Format

Please post all answers in the following format for the purpose of easy comparison:

## Language

Number of characters: ???

Fully obfuscated function:

``````(code here)
``````

Clear/semi-obfuscated function:

``````(code here)
``````

Any notes on the algorithm/clever shortcuts it takes.

+23  A:

## Python

Number of characters: 237

Fully obfuscated function:

``````from operator import*
def e(s,l=[]):
if s:l+=list(s.replace(' ','')+')')
while o:
c=p(0)
if c=='(':c=e(0)
while l[0]not in d:c+=p(0)
a=o(a,float(c));o=d[p(0)]
return a
``````

Clear/semi-obfuscated function:

``````import operator

def calc(source, stack=[]):
if source:
stack += list(source.replace(' ', '') + ')')

ops = {
')': 0,
'*': operator.mul,
'-': operator.sub,
'/': operator.div,
}

while op:
cur = stack.pop(0)

if cur == '(':
cur = calc(0)

while stack[0] not in ops:
cur += stack.pop(0)

op = ops[stack.pop(0)]

``````
Note: the current version doesn't handle an input like -(1.0)though it does handle negative literals correctly. It wasn't clear from the spec if this is required.
One can make l non-global for free by tucking it into the parameter list of e. Still won't be thread-safe, however.
Very cunning. That was well worth the effort of interpreting. :)
@Dave: Mine fails on `-(1.0)` too, so no worries! I'll clarify the question. Anyway, very clever solution it seems - I'm still trying to figure out how it works though (not exactly knowing Python). If you could add a brief explanation, that would be much appreciated.
+2  A:

## Ruby 1.9

(because of the regex)

Number of characters: 296

``````def d(s)
while m = s.match(/((?<pg>\((?:\\[()]|[^()]|\g<pg>)*\)))/)
s.sub!(m[:pg], d(m[:pg][1,m[:pg].size-2]))
end
while m = s.match(/(-?\d+(\.\d+)?)\s*([*+\-\/])\s*(-?\d+(\.\d+)?)/)
r=m[1].to_f.send(m[3],m[4].to_f) if %w{+ - * /}.include?m[3]
s.sub!(m[0], r.to_s)
end
s
end
``````

EDIT: Includes Martin's optimization.

r=m[1].to_f.send(m[3],m[4].to_f) if %w{+ - * /}.include?m[3]
Even better! I was trying to think of a nice way of doing that, and it skipped my mind.
+16  A:

## C99

Number of characters: 239 (But see below for 209)

compressed function:

``````#define S while(*e==32)++e
#define F float
F strtof();char*e;F v();F g(){S;return*e++-40?strtof(e-1,&e):v();}F v(){F b,a=g();for(;;){S;F o=*e++;if(!o|o==41)return a;b=g();a=o==43?a+b:o==45?a-b:o==42?a*b:a/b;}}F f(char*x){e=x;return v();}
``````

decompressed function:

``````float strtof();

char* e;
float v();

float g() {
while (*e == ' ') ++e;
return *e++ != '(' ? strtof(e-1, &e) : v();
}

float v() {
float b, a = g();
for (;;) {
while (*e == ' ') ++e;
float op = *e++;
if (op == 0 || op == ')') return a;
b = g();
a = op == '+' ? a + b : op == '-' ? a - b : op == '*' ? a * b : a / b;
}
}

float eval(char* x) {
e = x;
return v();
}
``````

Function is not re-entrant.

EDIT from Chris Lutz: I hate to trample on another man's code, but here is a 209-character version:

``````#define S for(;*e==32;e++)
#define X (*e++-40?strtof(e-1,&e):v())
float strtof();char*e;float v(){float o,a=X;for(;;){S;o=*e++;if(!o|o==41)return a;S;a=o-43?o-45?o-42?a/X:a*X:a-X:a+X;}}
#define f(x) (e=x,v())
``````

``````float strtof();
char *e;
float v() {
float o, a = *e++ != '(' ? strtof(e - 1, &e) : v();
for(;;) {
for(; *e == ' '; e++);
o = *e++;
if(o == 0 || o==')') return a;
for(; *e == ' '; e++);
// I have no idea how to properly indent nested conditionals
// and this is far too long to fit on one line.
a = o != '+' ?
o != '-' ?
o != '*' ?
a / (*e++ != '(' ? strtof(e - 1, &e) : v()) :
a * (*e++ != '(' ? strtof(e - 1, &e) : v()) :
a - (*e++ != '(' ? strtof(e - 1, &e) : v()) :
a + (*e++ != '(' ? strtof(e - 1, &e) : v());
}
}
#define f(x) (e = x, v())
``````

Yeah, `f()` is a macro, not a function, but it works. The readable version has some of the logic rewritten but not reordered (like `o != '+'` instead of `o - '+'`), but is otherwise just an indented (and preprocessed) version of the other one. I keep trying to simplify the `if(!o|o==41)return a;` part into the `for()` loop, but it never makes it shorter. I still believe it can be done, but I'm done golfing. If I work on this question anymore, it will be in the language that must not be named.

Nice solution, and bonus points for using "pure" C. Beats mine by 3 chars, as well! Re-entrancy wasn't in the rules, so that's fine. (It is a plus however.)
Nice! You can shave off a few more characters by using ASCII codes, e.g. replace '0' with 48, etc. And of course, you can save a bunch by using atof() instead of your home-grown float parser, but you're intentionally not using library functions, which is not a strict requirement of the problem.
I was thinking of using atof() but it doesn't tell you where the float string ends so you would need to parse it anyway.
Thanks for the tip, Adam. Using that and a couple of other (ugly) tricks, I shrunk it a bit further.
Ouch, I didn't count on negative numbers. Code inflated to 400 characters.
Oh right; I guess you can't use atof(), but you could use strtod() or sscanf() with a %n modifier.
One more microoptimization: replace "o==0||o==41" with "!o|o==41"
Nice tip concerning strtod(), I didn't even know about it. It cut it down quite a bit in spite of the fact that it needs a #include.
You can save two characters by using the fact that strtod() skips leading whitespace.
Not really, Dave, it still has to skip whitespace in the case where there is a ( and not a number.
Doh! I don't need the W macro anymore!
Yet another shaving: replace the #include with an exlpicit declaration of "double strtod();". In C (but NOT C++), an empty set of parens means the function takes any number of parameters of any type. Unfortunately, the declaration cannot be elided completely because it doesn't return int.
Actually, screw strtod() and use strtof() instead (C99 only)! Saves one more character (if you declare it yourself instead of including stdlib.h), and gets rid of a double-to-float warning.
Shaved a few more characters by making op a float instead of char. Made me a little queasy, but it worked.
Well done for the persistence! I'm not sure you're going to catch the Haskell/Python solutions, but you're amazingly close, much more so than I ever thought a solution in C would get. Interestingly, I believe this entire solution could be converted to C# at a minimal char cost.
You can save a few more characters by using something I used in my perl solution - the last octal digit of the character codes is different for each operator character, so {{F o=*e++;if(!o|o==41)...}} is equivalent to {{F o=7if(o<2)...}}. You should be able to squeeze some more characters out by using "<" and single digits for the rest of that long line; just sort by operator character code.
#define R return
@Nosredna: I thought of that but it actually wound up increasing the size by a couple of characters. One more return and it would have worked.
Oh yeah! You're right!
After a day of golfing, I've got one down to 214, but I'm going to see if I can shave off any more before I post it (I'm aiming to break 200, but I doubt it'll happen). Hints to what I did: g() and f() are simple enough to inline, strtof()'s chews up leading whitespace for you, o-41 is the same as o!=41 but shorter. I'll post code soon if I can't shave any more off.
209 and I'm posting it and freeing myself from this. Should I edit this post and add my solution or make a new one?
feel free to edit this post if you want
+ This is a riot. It is kinda like what I do in performance tuning big software, when I get the chance. I don't obfuscate, but I love lopping off big factors.
Woohoo! Clunky old C is now beating every Python solution up here, as well as Haskell!
Clever. Go team C!
Does the preprocessor let you do this and then use D for #define? If so, I can cut another couple characters, I think. --- #define D #define
@Nosredna - No, it doesn't. It checks macro expansions for other macros, but not for new preprocessor directives.
+2  A:

# Python

Number of characters: 492

Mildly obfuscated function (short variable names, no spaces around operators):

``````def e(s):
q=[]
b=1
v=[]
for c in s.replace(' ','')+'\$':
if c in '.0123456789' or c in '+-' and b and not v:
v+=[c]
else:
if v:
q+=[float(''.join(v))]
v=[]
while len(q)>=3:
x,y,z=q[-3:]
if type(x)==type(z)==float:
if y=='+':q[-3:]=[x+z]
elif y=='-':q[-3:]=[x-z]
elif y=='*':q[-3:]=[x*z]
elif y=='/':q[-3:]=[x/z]
elif (x,z)==('(',')'):q[-3:]=[y]
else:break
if c=='\$':break
q+=[c]
b=c!=')'
return q[0]
``````

I think this is relatively easy to understand. It's a pretty straightforward, naive approach. It doesn't import anything, doesn't use regex, is fully self-contained (single function, no globals, no side-effects), and should handle signed literals (positive or negative). Using more sensible variable names and adhering to recommended Python formatting increases the character count to more like 850-900, a big chunk of that from using four spaces instead of a single tab for indentation.

Does that char count include all the whitespace? If so, you could reduce it by a lot I'm sure. Well, the code pretty clear anyway, so good job with that. It's also a bonus that it's re-entrant.
My character count was what the editor told me was the size of the selection from the 'd' in "def" to the ']' at the end of the return. So that includes one character (a tab, for readability) for each indent level. This indentation is necessary for Python, so it has to be counted. Thanks for the comment; I knew I would never compete on pure character count or ingenuity.
Yeah, fair enough. I just observed that the other poster with a Python solution got away with using a single space for indentation, but if you're using tabs than that's equivalent.
I just noticed a brain fart in my code. Why did I make v a list instead of simply a string? The string not only saves 11 characters, but is also a little clearer, since it's what I eventually need anyway (Python's ''.join() isn't exactly the most intuitive).
+22  A:

Number of characters: 182

No attempt at cleverness, just some compression: 4 lines, 312 bytes.

``````import Data.Char;import Text.ParserCombinators.Parsec
q=either(error.show)id.runParser t id"".filter(' '/=);t=do
s<-getState;a<-fmap read(many1\$oneOf".-"<|>digit)<|>between(char '('>>setState id)(char ')'>>setState s)t
option(s a)\$choice(zipWith(\c o->char c>>return(o\$s a))"+-*/"[(+),(-),(*),(/)])>>=setState>>t
``````

And now, really getting into the golf spirit, 3 lines and 182 bytes:

``````q=snd.(`e`id).filter(' '/=)
e s c|[(f,h)]<-readsPrec 0 s=g h(c f);e('(':s)c=g h(c f)where(')':h,f)=e s id
g('+':h)=e h.(+);g('-':h)=e h.(-);g('*':h)=e h.(*);g('/':h)=e h.(/);g h=(,)h
``````

Exploded:

``````-- Strip spaces from the input, evaluate with empty accumulator,
-- and output the second field of the result.
q :: String -> Double
q = snd . flip eval id . filter (not . isSpace)

-- eval takes a string and an accumulator, and returns
-- the final value and what’s left unused from the string.
eval :: (Fractional a, Read a) => String -> (a -> a) -> (String, a)

-- If the beginning of the string parses as a number, add it to the accumulator,
-- then try to read an operator and further.
eval str accum | [(num, rest)] <- readsPrec 0 str = oper rest (accum num)

-- If the string starts parentheses, evaluate the inside with a fresh
-- accumulator, and continue after the closing paren.
eval ('(':str) accum = oper rest (accum num) where (')':rest, num) = eval str id

-- oper takes a string and current value, and tries to read an operator
-- to apply to the value.  If there is none, it’s okay.
oper :: (Fractional a, Read a) => String -> a -> (String, a)

-- Handle operations by giving eval a pre-seeded accumulator.
oper ('+':str) num = eval str (num +)
oper ('-':str) num = eval str (num -)
oper ('*':str) num = eval str (num *)
oper ('/':str) num = eval str (num /)

-- If there’s no operation parsable, just return.
oper str num = (str, num)
``````
I suspect that it may still be possible to get below 225, but this is as far as I can get at 3am.
That appears to be quite an elegant functional solution. (One that I certainly wouldn't have understand without the comments, so thanks for those.) Also, you're just marginally ahead of Dave's Python solution at the moment, so you seem to be leading! I'd be curious if an F# solution could match or even beat this.
It's interesting to me that the Parsec solution (parser combinators = generalized regex + more), even if minimization was attempted, doesn't come close to the hand-rolled parsing. I don't think F#'s syntax can be as concise as Haskell, but I'd welcome some competition too :)
@ephemient: I tried using both regex and the parser module in Python quickly. Was almost immediately below 300 chars, but saw no chance to get competitive. The issue is the import and the function calls eat up too much. That is true for most languages (exception Perl). BTW, you don't have to parse yourself to get substantially below 300 chars, as my solution shows.
I think you're over penalizing your character count. The problem asked for a String->Double function, so you should count characters by replacing "main=interact\$show." with "q=", for 17 more characters, putting your count at 209.
+10  A:

## JavaScript (Not IE compatible)

Number of characters: 268/260

Fully obfuscated function:

``````function e(x){x=x.replace(/ /g,'')+')'
function P(n){return x[0]=='('?(x=x.substr(1),E()):(n=/^[-+]?[\d.]+/(x)[0],x=x.substr(n.length),+n)}function E(a,o,b){a=P()
for(;;){o=x[0]
x=x.substr(1)
if(o==')')return a
b=P()
a=o=='+'?a+b:o=='-'?a-b:o=='*'?a*b:a/b}}return E()}
``````

or, in JavaScript 1.8 (Firefox 3+), you can save a few characters by using expression closures:

``````e=function(x,P,E)(x=x.replace(/ /g,'')+')',P=function(n)(x[0]=='('?(x=x.substr(1),E()):(n=/^[-+]?[\d.]+/(x)[0],x=x.substr(n.length),+n)),E=function(a,o,b){a=P()
for(;;){o=x[0]
x=x.substr(1)
if(o==')')return a
b=P()
a=o=='+'?a+b:o=='-'?a-b:o=='*'?a*b:a/b}},E())
``````

Clear/semi-obfuscated function:

``````function evaluate(x) {
x = x.replace(/ /g, "") + ")";
function primary() {
if (x[0] == '(') {
x = x.substr(1);
return expression();
}

var n = /^[-+]?\d*\.?\d*/.exec(x)[0];
x = x.substr(n.length);
return +n;
}

function expression() {
var a = primary();
for (;;) {
var operator = x[0];
x = x.substr(1);

if (operator == ')') {
return a;
}

var b = primary();
a = (operator == '+') ? a + b :
(operator == '-') ? a - b :
(operator == '*') ? a * b :
a / b;
}
}

return expression();
}
``````

Neither version will work in IE, because they use array-style subscripting on the string. If you replace both occurrences of `x[0]` with `x.charAt(0)`, the first one should work everywhere.

I cut out some more characters since the first version by turning variables into function parameters and replacing another if statement with the conditional operator.

That's pretty good. I was waiting for someone to use regex here. :) It would seem like dynamic languages definitely have an advantage for this problem.
+4  A:

## C#

Number of characters: 396 (updated)

(but fails the test you added with "/ -8", and I'm not inclined to fix it...

``````static float Eval(string s){int i,j;s=s.Trim();while((i=s.IndexOf(')'))>=0){j=s.LastIndexOf('(',i,i);s=s.Substring(0,j++)+Eval(s.Substring(j,i-j))+s.Substring(i+1);}if((i=s.LastIndexOfAny("+-*/".ToCharArray()))<0) return float.Parse(s);var r=float.Parse(s.Substring(i+1));var l=i>0?Eval(s.Substring(0,i)):(float?)null;return s[i]=='+'?(l??0)+r:(s[i]=='-'?(l??0)-r:(s[i]=='/'?(l??1)/r:(l??1)*r));}
``````

From:

``````static float Eval(string s)
{
int i, j;
s = s.Trim();
while ((i = s.IndexOf(')')) >= 0)
{
j = s.LastIndexOf('(', i, i);
s = s.Substring(0, j++) + Eval(s.Substring(j, i - j)) + s.Substring(i + 1);
}
if ((i = s.LastIndexOfAny("+-*/".ToCharArray())) < 0) return float.Parse(s);
var r = float.Parse(s.Substring(i + 1));
var l = i > 0 ? Eval(s.Substring(0, i)) : (float?)null;
return s[i] == '+'
? (l ?? 0) + r
: (s[i] == '-'
? (l ?? 0) - r
: (s[i] == '/'
? (l ?? 1) / r
: (l ?? 1) * r));
}
``````
Ah wonderful, a C# solution. Your use of nullable types in particular is quite interesting. 484 seems pretty good, given that you didn't have the time to tidy it up. (One improvement would be to convert the switch statement into a series of ifs, I believe.) I've posted my own C# solution now, if you wish to compare. :)
+5  A:

## C#

Number of characters: 403

So here's my solution... I'm still waiting for someone to post one in C# that can beat it. (Marc Gravell was close, and may yet do better than me after some more tinkering.)

Fully obfuscated function:

``````float e(string x){float v=0;if(float.TryParse(x,out v))return v;x+=';';int t=0;
char o,s='?',p='+';float n=0;int l=0;for(int i=0;i<x.Length;i++){o=s;if(
x[i]!=' '){s=x[i];if(char.IsDigit(x[i])|s=='.'|(s=='-'&o!='1'))s='1';if(s==')')
l--;if(s!=o&l==0){if(o=='1'|o==')'){n=e(x.Substring(t,i-t));if(p=='+')v+=n;
if(p=='-')v-=n;if(p=='*')v*=n;if(p=='/')v/=n;p=x[i];}t=i;if(s=='(')t++;}
if(s=='(')l++;}}return v;}
``````

Semi-obfuscated function:

``````public static float Eval(string expr)
{
float val = 0;
if (float.TryParse(expr, out val))
return val;
expr += ';';
int tokenStart = 0;
char oldState, state = '?', op = '+';
float num = 0;
int level = 0;
for (int i = 0; i < expr.Length; i++)
{
oldState = state;
if (expr[i] != ' ')
{
state = expr[i];
if (char.IsDigit(expr[i]) || state == '.' ||
(state == '-' && oldState != '1'))
state = '1';
if (state == ')')
level--;
if (state != oldState && level == 0)
{
if (oldState == '1' || oldState == ')')
{
num = Eval(expr.Substring(tokenStart, i - tokenStart));
if (op == '+') val += num;
if (op == '-') val -= num;
if (op == '*') val *= num;
if (op == '/') val /= num;
op = expr[i];
}
tokenStart = i;
if (state == '(')
tokenStart++;
}
if (state == '(')
level++;
}
}
return val;
}
``````

Nothing too clever going on here, it woul seem. The function does however have the advantage of being re-entrant (i.e. thread-safe).

I am also reasonably pleased with the number of chars, given that it's written in C# (valid 1.0, 2.0, and 3.0 I believe).

Any tips on how I might reduce the char count further would be welcome. (The is my first real attempt at code golf.)
I got it < 400, but it fails the edited test you added ;-p
Suggestions: "var" for float, char - only shaves a few, and loses the C# 1.2/2.0 compatibility, though.
@Marc: Yeah, that's about as far as I got too. With a few other minor changes, I might get it down to 390, but no less.
Nice solution Nolorin. I was able to get your solution down to 361
@Chris: Thanks - though now I'm curious how you managed to get the char count down so much. (I could probably knock off about 15 or 20 more myself, but no more.) Mind sharing it with us? Feel free to edit my post, even.
I've gotten it down to 294 chars using a bit of regex. :) See my new answer: http://stackoverflow.com/questions/928563/code-golf-evaluating-mathematical-expressions/944716#944716
Doesn't work for unary minus: (5 + 5) * (3 + 2)-1 Evaluates to 51.
+3  A:

## Ruby 1.8.7

Number of characters: 620

Do try and take it easy on my implementation, it's the first time I've written an expression parser in my life! I guarantee that it isn't the best.

Obfuscated:

``````def solve_expression(e)
t,r,s,c,n=e.chars.to_a,[],'','',''
while(c=t.shift)
n=t[0]
if (s+c).match(/^(-?)[.\d]+\$/) || (!n.nil? && n.match(/\d/) && c=='-')
s+=c
elsif (c=='-' && n=='(') || c=='('
m,o,x=c=='-',1,''
while(c=t.shift)
o+=1 if c=='('
o-=1 if c==')'
x+=c unless c==')' && o==0
break if o==0
end
r.push(m ? -solve_expression(x) : solve_expression(x))
s=''
elsif c.match(/[+\-\/*]/)
r.push(c) and s=''
else
r.push(s) if !s.empty?
s=''
end
end
r.push(s) unless s.empty?
i=1
a=r[0].to_f
while i<r.count
b,c=r[i..i+1]
c=c.to_f
case b
when '+': a=a+c
when '-': a=a-c
when '*': a=a*c
when '/': a=a/c
end
i+=2
end
a
end
``````

``````def solve_expression(expr)
chars = expr.chars.to_a # characters of the expression
parts = [] # resulting parts
s,c,n = '','','' # current string, character, next character

while(c = chars.shift)
n = chars[0]
if (s + c).match(/^(-?)[.\d]+\$/) || (!n.nil? && n.match(/\d/) && c == '-') # only concatenate when it is part of a valid number
s += c
elsif (c == '-' && n == '(') || c == '(' # begin a sub-expression
negate = c == '-'
open = 1
subExpr = ''
while(c = chars.shift)
open += 1 if c == '('
open -= 1 if c == ')'
# if the number of open parenthesis equals 0, we've run to the end of the
# expression.  Make a new expression with the new string, and add it to the
# stack.
subExpr += c unless c == ')' && open == 0
break if open == 0
end
parts.push(negate ? -solve_expression(subExpr) : solve_expression(subExpr))
s = ''
elsif c.match(/[+\-\/*]/)
parts.push(c) and s = ''
else
parts.push(s) if !s.empty?
s = ''
end
end
parts.push(s) unless s.empty? # expression exits 1 character too soon.

# now for some solutions!
i = 1
a = parts[0].to_f # left-most value is will become the result
while i < parts.count
b,c = parts[i..i+1]
c = c.to_f
case b
when '+': a = a + c
when '-': a = a - c
when '*': a = a * c
when '/': a = a / c
end
i += 2
end
a
end
``````
That's quite good for a first attempt, and the length isn't hugely off the others anyway. Certainly, the algorithm is pretty clear. Note that you can reduce the char count significantly just by using one-letter variable names!
Thanks. My last bug took awhile to fix, but in general it wasn't anything brain-wracking; thankfully it functions fully.
+6  A:

## PHP

Number of characters: 284

obfuscated:

``````function f(\$m){return c(\$m[1]);}function g(\$n,\$m){\$o=\$m[0];\$m[0]=' ';return\$o=='+'?\$n+\$m:(\$o=='-'?\$n-\$m:(\$o=='*'?\$n*\$m:\$n/\$m));}function c(\$s){while(\$s!=(\$t=preg_replace_callback('/\(([^()]*)\)/',f,\$s)))\$s=\$t;preg_match_all('![-+/*].*?[\d.]+!',"+\$s",\$m);return array_reduce(\$m[0],g);}
``````

``````function callback1(\$m) {return c(\$m[1]);}
function callback2(\$n,\$m) {
\$o=\$m[0];
\$m[0]=' ';
return \$o=='+' ? \$n+\$m : (\$o=='-' ? \$n-\$m : (\$o=='*' ? \$n*\$m : \$n/\$m));
}
function c(\$s){
while (\$s != (\$t = preg_replace_callback('/\(([^()]*)\)/','callback1',\$s))) \$s=\$t;
preg_match_all('![-+/*].*?[\d.]+!', "+\$s", \$m);
return array_reduce(\$m[0], 'callback2');
}

\$str = '  2.45/8.5  *  -9.27   +    (   5   *  0.0023  ) ';
var_dump(c(\$str));
# float(-2.66044117647)
``````

Should work with any valid input (including negative numbers and arbitrary whitespace)

+12  A:

## Common Lisp

(SBCL)
Number of characters: 251

``````(defun g(e)(if(numberp e)e(let((m (g (pop e)))(o(loop for x in e by #'cddr collect x))(n(loop for x in (cdr e)by #'cddr collect (g x))))(mapcar(lambda(x y)(setf m(apply x(list m y))))o n)m)))(defun w(e)(g(read-from-string(concatenate'string"("e")"))))
``````

Proper version (387 chars):

``````(defun wrapper (exp) (golf-eval (read-from-string (concatenate 'string "(" exp ")"))))

(defun golf-eval (exp)
(if (numberp exp)
exp
(let ((mem (golf-eval (pop exp)))
(op-list (loop for x in exp by #'cddr collect x))
(num-list (loop for x in (cdr exp) by #'cddr collect (golf-eval x))))
(mapcar (lambda (x y) (setf mem (apply x (list mem y)))) op-list num-list)
mem)))
``````

Input is form `w()`, which takes one string argument. It uses the trick that nums/operands and operators are in the pattern N O N O N ... and recursively evaluates all operands, and therefore getting nesting very cheap. ;)

Clever solution. Nonetheless, I'm not quite sure it's completely valid given that the specification was for the function to take a string object.
No problem. Didn't realise the conversion was so easy. Good solution, still!
Wow. That's beutiful. :)
+2  A:

## Python 3K

(its 3K because / converts the result to a floating point number)

Number of characters: 808

Clear (I cannot write obfuscated code in Python XD):

``````def parse(line):
ops = {"+": lambda x,y:x+y,
"-": lambda x,y:x-y,
"*": lambda x,y:x*y,
"/": lambda x,y:x/y}
def tpp(s, t):
if len(s) > 0 and s[-1] in ops:
f = ops[s.pop()]
t = f(s.pop(), t)
return t
line = line + " "
s = []
t = 0
m = None
for c in line:
if c in "0123456789":
if not m:
m = "i"
if m == "i":
t = t*10 + ord(c)-ord("0")
elif m =="d":
t = t + e*(ord(c)-ord("0"))
e*=0.1
elif c == ".":
m = "d"
e = 0.1
elif m:
t = tpp(s,t)
s.append(t)
m = None
t = 0

if c in ops or c == "(":
s.append(c)
elif c == ")":
t = s.pop()
s.pop()
s.append(tpp(s,t))
t = 0
t = s.pop()
if int(t) == t:
t = int(t)
return t
``````

I'm not using any kind of regular expression, even the number parsing is made by hand ;-)

Quite simple, scans the line, it can be in 3 different modes (m), None that means that there's no number being parsed, "i" that means that it is parsing the integer part and "d" that means that is parsing the decimal part.

It uses a stack to store the temporary computations, when it has finished parsing a number sees if it there was an operator in the stack, in that case evals and pushes. The opening parens are just pushed and the closing parens remove the opening paren and repush the current eval.

Fairly simple and straightfordward :-)

+6  A:

## F#

Number of characters: 327

OP was looking for an F# version, here it is. Can be done a lot nicer since I'm abusing a ref here to save characters. It handles most things such as -(1.0), 3 - -3 and even 0 - .5 etc.

``````let g s=
let c=ref[for x in System.Text.RegularExpressions.Regex.Matches(s,"[0-9.]+|[^\s]")->x.Value]
let rec e v=if (!c).IsEmpty then v else
c:=(!c).Tail
match h with|"("->e(e 0.0)|")"->v|"+"->e(v+(e 0.0))|"-"->e(v-(e 0.0))|"/"->e(v/(e 0.0))|"*"->e(v*(e 0.0))|x->float x
e(e 0.0)
``````
Indeed, I was hoping for an F# solution. Thanks for that. Char count is pretty decent too, especially considering that "System.Text.RegularExpressions.Regex.Matches" takes up an absurd number of characters.
yeah, same with the .Value.IsEmpty/Tail/Head calls - I got a new version in the works ;p hoping for sub250 chars.
I'm not actually sure whether in some code golf contests you are allowed import/using statements outside the char count. That would definitely help, if so. :) Looking forward to seeing the new version.
@Noldorin: Nope I'm sorry I can't get it under the 327 characters of this (a bit enhanced since last) code. The gain from having everything perfectly parsed with the regex outweighs the insanely long name of "System.Text.RegularExpressions.Regex.Matches"If F# would've had a short (aliased) name for the Matches function I would be at 288 chars, but it does not =/.
@fredrikholmstrom: No worries - good solution nonetheless. Also, I'm not completely sure, but I would say that you should be able to move "System.Text.RegularExpressions" into an "open" statement and exclude the char count for that at least.
+2  A:

## Ruby

Number of characters: 302

Semi-obfuscated:

``````def e(l)
t=0.0;o=nil
while l!=''
l.sub!(/^\s+/,'')
l.sub!(/^(-?\d+|-?\d+\.\d+)/,'')
t=o ? t.send(o, \$1.to_f) : \$1.to_f if \$~
l.sub!(/^(\+|-|\*|\/)/,'')
o=\$1 if \$~
l.sub!(/^\(/,'')
t=o ? t.send(o, e(l)) : e(l) if \$~
l.sub!(/^\)/,'')
return t if \$~
end
t
end
``````

Destroys original string, also assumes expression is well-formed (only valid characters, and matching brackets).

Not obfuscated:

``````def evaluate_expression(expression)
result_so_far = 0.0
last_operator = nil

while (expression != '')
expression.sub!(/^\s+/, '')

# extract and remove leading integer or decimal number
expression.sub!(/^(-?\d+|-?\d+\.\d+)/, '')
if \$~
# match was successful
number = \$1.to_f
if last_operator.nil?
# first number, just store it
result_so_far = number
else
# we have an operator, use it!
# last_operator is a string matching '+', '-', '*' or '/'
# just invoke the method of that name on our result_so_far
# since these operators are just method calls in Ruby
result_so_far = result_so_far.send(last_operator, number)
end
end

# extract and remove leading operator +-*/
expression.sub!(/^(\+|-|\*|\/)/, '')
if \$~
# match was successful
last_operator = \$1
end

# extract and remove leading open bracket
l.sub!(/^\(/, '')
if \$~
# match successful
if last_operator.nil?
# first element in the expression is an open bracket
# so just evaluate its contents recursively
result_so_far = evaluate_expression(expression)
else
# combine the content of the bracketing with the
# result so far using the last_operator
result_so_far.send(last_operator, evaluate_expression(expression))
end
end

# extract and remove leading close bracket
l.sub!(/^\)/, '')
if \$~
# match successful
# this must be the end of a recursive call so
# return the result so far without consuming the rest
# of the expression
return result_so_far
end
end
t
end
``````

The recursive call is controlled by the modification of the expression string, which is a bit nasty, but it seems to work.

Could you put in a readable version as well? I'm not entirely sure how yours works.
+3  A:

## Python with regular expressions

Number of characters: 283

Fully obfuscated function:

``````import re
from operator import*
def c(e):
for v,o in re.findall("(-?[.\d]+)|([+-/*()])",e):
if v:s=[float(v)]+s
elif o=="(":s=a+s
elif o!=")":s=[O[o]]+s
if v or o==")":s[:3]=[s[1](s[2],s[0])]
return s[0]
``````

Not obfuscated:

``````import re
from operator import *

def compute(s):
operators = dict(zip("+-/*()", (add, sub, truediv, mul)))
for val, op in re.findall("(-?[.\d]+)|([+-/*()])", s):
if val:
stack = [float(val)] + stack
elif op == "(":
stack = [add, 0] + stack
elif op != ")":
stack = [operators[op]] + stack
if val or op == ")":
stack[:3] = [stack[1](stack[2], stack[0])]
return stack[0]
``````

I wanted to see if I cab beat the other Python solutions using regular expressions.

Couldn't.

The regular expression I'm using creates a list of pairs (val, op) where only one item in each pair is valid. The rest of the code is a rather standard stack based parser with a neat trick of replacing the top 3 cells in the stack with the result of the computation using Python list assignment syntax. Making this work with negative numbers required only two additional characters (-? in the regex).

You can save a couple of bytes by removing "()" from your operator string; `zip` stops at the end of the shorter list.
@gooli: Are you using Windows? By my count, the posted solution is only 273. One explanation for this may be that you counted newlines as two characters each. (Python doesn't care if you have single-char newlines, even in Windows.) Another explanation is that hit 8 when you meant 7. ;)
+7  A:

## C# with Regex Love

Number of characters: 384

Fully-obfuscated:

``````float E(string i){i=i.Replace(" ","");Regex b=new Regex(@"\((?>[^()]+|\((?<D>)|\)(?<-D>))*(?(D)(?!))\)");i=b.Replace(i,m=>Eval(m.Value.Substring(1,m.Length-2)).ToString());float r=0;foreach(Match m in Regex.Matches(i,@"(?<=^|\D)-?[\d.]+")){float f=float.Parse(m.Value);if(m.Index==0)r=f;else{char o=i[m.Index-1];if(o=='+')r+=f;if(o=='-')r-=f;if(o=='*')r*=f;if(o=='/')r/=f;}}return r;}
``````

Not-obfuscated:

``````private static float Eval(string input)
{
input = input.Replace(" ", "");
Regex balancedMatcher = new Regex(@"\(
(?>
[^()]+
|
\( (?<Depth>)
|
\) (?<-Depth>)
)*
(?(Depth)(?!))
\)", RegexOptions.IgnorePatternWhitespace);
input = balancedMatcher.Replace(input, m => Eval(m.Value.Substring(1, m.Length - 2)).ToString());

float result = 0;

foreach (Match m in Regex.Matches(input, @"(?<=^|\D)-?[\d.]+"))
{
float floatVal = float.Parse(m.Value);
if (m.Index == 0)
{
result = floatVal;
}
else
{
char op = input[m.Index - 1];
if (op == '+') result += floatVal;
if (op == '-') result -= floatVal;
if (op == '*') result *= floatVal;
if (op == '/') result /= floatVal;
}
}

return result;
}
``````

Takes advantage of .NET's Regex balancing group feature.

Thanks for that solution. :) I wasn't sure whether I would see a C# solution with regex, but here we have it. Now, it's arguable whether you should include the "using System.Text.RegularExpressions;" in your char count, but it's a good solution nonetheless.
That wasn't part of the rules :). If you add "using R=System.Text.RegularExpressions.Regex;" and replace my "Regex" with R, it goes to 417.
@Jeff: Well, technically it won't compile without the using statement, so by default it *should* be included. Petty point, however, given that our C# solutions are all significantly behind the leader.
Agreed. Just trying to have regex fun.
Yeah, regex for the win! :)
+1  A:

## F#

Number of characters: 461

Here is Marc Gravell's solution (essentially) converted from C# to F#. The char count is scarecly better, but I thought I'd post it anyway out of interest.

Obfuscated code:

``````let e x=
let rec f(s:string)=
let i=s.IndexOf(')')
if i>0 then
let j=s.LastIndexOf('(',i)
f(s.Substring(0,j)+f(s.Substring(j+1,i-j-1))+s.Substring(i+1))
else
let o=[|'+';'-';'*';'/'|]
let i=s.LastIndexOfAny(o)
let j=s.IndexOfAny(o,max(i-2)0,2)
let k=if j<0 then i else j
if k<0 then s else
let o=s.[k]
string((if o='+'then(+)else if o='-'then(-)else if o='*'then(*)else(/))(float(f(s.Substring(0,k))))(float(s.Substring(k+1))))
float(f x)
``````
Yeah I tried something similar with F# yesterday but it both feels very "non F#" and is still more chars then my other solution ;(. Silly that F# has so horrible string handling.
+3  A:

## Python

Number of characters: 382

Yet another Python solution, heavily using regular expression replacement. Each run through the loop the simplest expressions are computed and the results are put back into the string.

This is the unobfuscated code, unless you consider regular expressions to be obfuscated.

``````import re
from operator import *
operators = dict(zip("+-/*", (add, sub, truediv, mul)))
def compute(s):
def repl(m):
v1, op, v2 = m.groups()
return str(operators[op](float(v1), float(v2)))
while not re.match("^\d+\.\d+\$", s):
s = re.sub("([.\d]+)\s*([+-/*])\s*([.\d]+)", repl, s)
s = re.sub("\(([.\d]+)\)", r"\1", s)
return s
``````

Had this idea just as I was turning in and couldn't let it go until I wrote it down and made it work.

Nice solution... Looks very clear to me, too. It would seem that using dict/zip to store the operators is definitely a very effective approach in Python.
A:

## Perl

Number of characters: 93

Fully obfuscated function: (93 characters if you join these three lines into one)

``````\$_="(@ARGV)";s/\s//g;\$n=qr/(-?\d+(\.\d+)?)/;
while(s.\(\$n\)|(?<=\()\$n[-+*/]\$n.eval\$&.e){}
print
``````

Clear/semi-obfuscated function:

``````\$_="(@ARGV)";            # Set the default var to "(" argument ")"
s/\s//g;                 # Strip all spaces from \$_
\$n=qr/(-?\d+(\.\d+)?)/;  # Compile a regex for "number"

# repeatedly replace the sequence "(" NUM ")" with NUM, or if there aren't
# any of those, replace "(" NUM OP NUM with the result
# of doing an eval on just the NUM OP NUM bit.
while(s{\(\$n\)|(?<=\()\$n[-+*/]\$n}{eval\$&}e){}

# print \$_
print
``````

I think this is pretty well explained in the "clear" version. The two main insights are that you can make the code uniform by surrounding the argument with parentheses at the start (special cases cost characters), and that it is sufficient, albeit massively inefficient, to only process stuff right next to an open parenthesis, replacing it with its result.

It's probably easiest to run this code as:

``````perl -le '\$_="(@ARGV)";s/\s//g;\$n=qr/(-?\d+(\.\d+)?)/;while(s.\(\$n\)|(?<=\()\$n[-+*/]\$n.eval\$&.e){}print' '4 * (9 - 4) / (2 * 6 - 2) + 8'
``````
Interesting, but does the "eval" kinda break the rules?
Ah, okay, it wasn't clear to me that "eval" was disallowed; I thought the description was saying that his "no BODMAS" rule was sufficient guard against that. Very well, I'll try another solution without 'eval'.
Yeah, sorry. The rules were to explicitly disallow "eval". It was just a suspicion that BODMAS *ought* to prevent against it, but not necessarily.
+18  A:

## Fortran 77 (gfortran dialect, now with g77 support)

Number of characters: 2059

Obfuscated version:

``````      function e(c)
character*99 c
character b
real f(24)
integer i(24)
nf=0
ni=0
20   nf=kf(0.0,nf,f)
ni=ki(43,ni,i)
30   if (isp(c).eq.1) goto 20
h=fr(c)
31   g=fp(nf,f)
j=ip(ni,i)
select case(j)
case (40)
goto 20
case (42)
d=g*h
case (43)
d=g+h
case (45)
d=g-h
case (47)
d=g/h
end select
50   nf=kf(d,nf,f)
60   j=nop(c)
goto (20, 70, 75, 75, 60, 75, 60, 75) (j-39)
65   e=fp(nf,f)
return
70   h=fp(nf,f)
goto 31
75   ni=ki(j,ni,i)
goto 30
end
function kf(v,n,f)
real f(24)
kf=n+1
f(n+1)=v
return
end
function ki(j,n,i)
integer i(24)
ki=n+1
i(n+1)=j
return
end
function fp(n,f)
real f(24)
fp=f(n)
n=n-1
return
end
function ip(n,i)
integer i(24)
ip=i(n)
n=n-1
return
end
function nop(s)
character*99 s
l=1
do while(s(l:l).eq." ".and.l.lt.99)
l=l+1
enddo
nop=ichar(s(l:l))
s(l:l)=" "
return
end
function isp(s)
character*99 s
isp=0
l=1
do while(s(l:l).eq." ".and.l.lt.99)
l=l+1
enddo
isp=41-ichar(s(l:l))
if (isp.eq.1) s(l:l)=" "
return
end
function fr(s)
character*99 s
m=1
n=1
i=1
do while(i.le.99)
j=ichar(s(i:i))
if (j.eq.32) goto 90
if (j.ge.48.and.j.lt.58) goto 89
if (j.eq.43.or.j.eq.45) goto (89,80) m
if (j.eq.46) goto (83,80) n
80      exit
83      n=2
89      m=2
90      i=i+1
enddo
do 91 j=1,i-1
s(j:j)=" "
91   continue
return
end
``````

Clear version: (3340 characters with scaffold)

``````      program infixeval
character*99 c
do while (.true.)
do 10 i=1,99
c(i:i)=" "
10      continue
f=e(c)
write(*,*)f
enddo
end

function e(c)
character*99 c
character b
real f(24)                ! value stack
integer i(24)             ! operator stack
nf=0                      ! number of items on the value stack
ni=0                      ! number of items on the operator stack
20   nf=pushf(0.0,nf,f)
ni=pushi(43,ni,i)         ! ichar(+) = 43
D     write (*,*) "'",c,"'"
30   if (isp(c).eq.1) goto 20
h=fr(c)
D     write (*,*) "'",c,"'"
31   g=fpop(nf,f)
j=ipop(ni,i)
D     write(*,*) "Opperate ",g," ",char(j)," ",h
select case(j)
case (40)
goto 20
case (42)                 ! "*"
d=g*h
case (43)                 ! "+"
d=g+h
case (45)                 ! "-"
d=g-h
case (47)                 ! "*"
d=g/h
end select
50   nf=pushf(d,nf,f)
60   j=nop(c)
D     write(*,*) "Got op: ", char(j)
goto (20, 70, 75, 75, 60, 75, 60, 75) (j-39)
65   e=fpop(nf,f)
return
70   h=fpop(nf,f)              ! Encountered a "("
goto 31
75   ni=pushi(j,ni,i)
goto 30
end

c     push onto a real stack
c     OB as kf
function pushf(v,n,f)
real f(24)
pushf=n+1
f(n+1)=v
D     write(*,*) "Push ", v
return
end

c     push onto a integer stack
c     OB as ki
function pushi(j,n,i)
integer i(24)
pushi=n+1
i(n+1)=j
D     write(*,*) "Push ", char(j)
return
end

c     pop from real stack
c     OB as fp
function fpop(n,f)
real f(24)
fpop=f(n)
n=n-1
D      write (*,*) "Pop ", fpop
return
end

c     pop from integer stack
c     OB as ip
function ipop(n,i)
integer i(24)
ipop=i(n)
n=n-1
D      write (*,*) "Pop ", char(ipop)
return
end

c     Next OPerator: returns the next nonws character, and removes it
c     from the string
function nop(s)
character*99 s
l=1
do while(s(l:l).eq." ".and.l.lt.99)
l=l+1
enddo
nop=ichar(s(l:l))
s(l:l)=" "
return
end

c     IS an open Paren: return 1 if the next non-ws character is "("
c     (also overwrite it with a space. Otherwise return not 1
function isp(s)
character*99 s
isp=0
l=1
do while(s(l:l).eq." ".and.l.lt.99)
l=l+1
enddo
isp=41-ichar(s(l:l))
if (isp.eq.1) s(l:l)=" "
return
end

c     Float Read: return the next real number in the string and removes the
c     character
function fr(s)
character*99 s
m=1                      ! No sign (Minus or plus) so far
n=1                      ! No decimal so far
i=1
do while(i.le.99)
j=ichar(s(i:i))
if (j.eq.32) goto 90   ! skip spaces
if (j.ge.48.and.j.lt.58) goto 89
if (j.eq.43.or.j.eq.45) goto (89,80) m
if (j.eq.46) goto (83,80) n
c     not part of a number
80      exit
83      n=2
89      m=2
90      i=i+1
enddo
do 91 j=1,i-1
s(j:j)=" "
91   continue
return
end
``````

Notes This edited version is rather more evil than my first attempt. Same algorithm, but now inline with a horrible tangle of `goto`s. I've ditched the co-routines, but am now using a couple of flavors of computed branches. All error checking and reporting has been removed, but this version will silently recover from some classes of unexpected characters in the input. This version also compiles with g77.

The primary limits are still fortran's rigid formatting, long and ubiquitous keywords, and simple primitives.

Good God, man! You must have been bored today. ;)
Hehe, I don't think I was ever expecting a Fortran solution! I think we can conclude that the language isn't particularly well-suited to code golf? Up-voted anyway for the sheer effort and for using an antiquated language. :)
I find this kind of fiddly byte diddling to be wordy and awkward in fortran, but not actually hard. Writing unstructured code and using those computed branches, on the other hand, feels kinda kinky.
Nicely done, but how does a 2000+ character fortran version get more votes than my short little ruby1.9 version? lol
@darkhelmet: I have no idea. I did it on a lark and expected one or two votes for effort and perversity. I am obscenely proud of this abomination, but this is ridiculous...
+31  A:

## Perl (no eval)

Number of characters: 167 106 (see below for the 106 character version)

Fully obfuscated function: (167 characters if you join these three lines into one)

``````sub e{my\$_="(\$_[0])";s/\s//g;\$n=q"(-?\d++(\.\d+)?+)";
@a=(sub{\$1},1,sub{\$3*\$6},sub{\$3+\$6},4,sub{\$3-\$6},6,sub{\$3/\$6});
while(s:\(\$n\)|(?<=\()\$n(.)\$n:\$a[7&ord\$5]():e){}\$_}
``````

Clear/deobfuscated version:

``````sub e {
my \$_ = "(\$_[0])";
s/\s//g;
\$n=q"(-?\d++(\.\d+)?+)"; # a regex for "number", including capturing groups
# q"foo" in perl means the same as 'foo'
# Note the use of ++ and ?+ to tell perl
# "no backtracking"

@a=(sub{\$1},             # 0 - no operator found
1,                   # placeholder
sub{\$3*\$6},          # 2 - ord('*') = 052
sub{\$3+\$6},          # 3 - ord('+') = 053
4,                   # placeholder
sub{\$3-\$6},          # 5 - ord('-') = 055
6,                   # placeholder
sub{\$3/\$6});         # 7 - ord('/') = 057

# The (?<=... bit means "find a NUM WHATEVER NUM sequence that happens
# immediately after a left paren", without including the left
# paren.  The while loop repeatedly replaces "(" NUM WHATEVER NUM with
# "(" RESULT and "(" NUM ")" with NUM.  The while loop keeps going
# so long as those replacements can be made.

while(s:\(\$n\)|(?<=\()\$n(.)\$n:\$a[7&ord\$5]():e){}

# A perl function returns the value of the last statement
\$_
}
``````

I had misread the rules initially, so I'd submitted a version with "eval". Here's a version without it.

The latest bit of insight came when I realized that the last octal digit in the character codes for `+`, `-`, `/`, and `*` is different, and that `ord(undef)` is 0. This lets me set up the dispatch table `@a` as an array, and just invoke the code at the location `7 & ord(\$3)`.

There's an obvious spot to shave off one more character - change `q""` into `''` - but that would make it harder to cut-and-paste into the shell.

## Even shorter

Number of characters: 124 106

Taking edits by ephemient into account, it's now down to 124 characters: (join the two lines into one)

``````sub e{\$_=\$_[0];s/\s//g;\$n=q"(-?\d++(\.\d+)?+)";
1while s:\(\$n\)|\$n(.)\$n:(\$1,1,\$3*\$6,\$3+\$6,4,\$3-\$6,6,\$6&&\$3/\$6)[7&ord\$5]:e;\$_}
``````

## Shorter still

Number of characters: 110 106

The ruby solution down below is pushing me further, though I can't reach its 104 characters:

``````sub e{(\$_)[email protected]_;\$n='( *-?[.\d]++ *)';
s:\(\$n\)|\$n(.)\$n:((\$1,\$2-\$4,\$4&&\$2/\$4,\$2*\$4,\$2+\$4)x9)[.8*ord\$3]:e?e(\$_):\$_}
``````

I had to give in and use `''`. That ruby `send` trick is really useful for this problem.

## Squeezing water from a stone

Number of characters: 106

A small contortion to avoid the divide-by-zero check.

``````sub e{(\$_)[email protected]_;\$n='( *-?[.\d]++ *)';
s:\(\$n\)|\$n(.)\$n:(\$1,0,\$2*\$4,\$2+\$4,0,\$2-\$4)[7&ord\$3]//\$2/\$4:e?e(\$_):\$_}
``````

Here's the test harness for this function:

``````perl -le 'sub e{(\$_)[email protected]_;\$n='\''( *-?[.\d]++ *)'\'';s:\(\$n\)|\$n(.)\$n:(\$1,0,\$2*\$4,\$2+\$4,0,\$2-\$4)[7&ord\$3]//\$2/\$4:e?e(\$_):\$_}' -e 'print e(\$_) for @ARGV' '1 + 3' '1 + ((123 * 3 - 69) / 100)' '4 * (9 - 4) / (2 * 6 - 2) + 8' '2*3*4*5+99' '2.45/8.5*9.27+(5*0.0023) ' '1 + 3 / -8'
``````
That's craziness. I suppose the the original code golf language ought to have beaten them all here, and indeed it has. I can't say I much like the abundance of regex, but good job! I might just have to accept this as answer...
thats crafty using ord. I like how you've used \$n as a way to duplicate a section of the regular expression.
Is there a reason you use "local" when "my" should work just as well?
@Chris Lutz: No, so I changed it and lost three more characters. Just the force of powerful habit telling me to use "local" with global variables.
Ugh... in my edit, qw// was supposed to be qr//. Since that (accidentally) works, you could shave another 2 characters off by replacing it with ''.
In fact, ephemient, I was already doing that (using q""). But nice trick to avoid all the "sub" noise.
That's impressive - you shaved it down AND added in divide-by-zero protection. I was going to suggest using "1 while" instead of "while(){}", but I didn't know that 1while (without the space) would parse correctly, so it didn't save any space. Glad it works.
It's pretty scary how small Perl can go, I edited my answer to keep it the smallest Ruby implementation and ran out of room at 170 chars. But 124? Good gravy!
I didn't notice that nobody has mentioned it yet, but this solution requires Perl 5.10. For compatibility with 5.8, use (-?(?>\d+(\.\d+)?)) which is two characters longer.
perl. is. sick.
@Epaga, don't worry I got your typo: perl. is. awesome.
Shorten it by 1 character - change "\$_=\$_[0]" to "(\$_)[email protected]_".
Do you really need to check for division by zero? Most of the solutions (including mine) do not.
Because it unconditionally performs the arithmetic regardless of the operator (picking the correct result later), it does need to avoid dividing by zero.
Used another Perl 5.10 trick to cut the character count a little more, but geez, this is getting really tough. Getting to 103 will not be easy.
Dare I tempt you to break 100? :P No really, please don't spend any more time on it unless you absolutely want to (for yourself)! It's a great solution as it is - certainly good enough for me.
It's more like a stubborn refusal to believe that Perl could be permanently beaten by its younger cousin Ruby, but yeah, it's not really a productive use of time...
I found a way to make it shorter, but it doesn't work. Negative array indices in Perl wrap once, but if they wrap past the end of the array, they don't keep wrapping. Otherwise, I could have had it to 105. (It lined up perfectly, too, which is really bothering me - if it kept wrapping, element -42 would have ended up as index 2, and it would have worked perfectly.)
+4  A:

## Python

Number of characters: 235

Fully obfuscated function:

``````def g(a):
i=len(a)
while i:
try:m=g(a[i+1:]);n=g(a[:i]);a=str({'+':n+m,'-':n-m,'*':n*m,'/':n/(m or 1)}[a[i]])
except:i-=1;j=a.rfind('(')+1
if j:k=a.find(')',j);a=a[:j-1]+str(g(a[j:k]))+a[k+1:]
return float(a.replace('--',''))
``````

Semi-obfuscated:

``````def g(a):
i=len(a);
# do the math
while i:
try:
# recursively evaluate left and right
m=g(a[i+1:])
n=g(a[:i])
# try to do the math assuming that a[i] is an operator
a=str({'+':n+m,'-':n-m,'*':n*m,'/':n/(m or 1)}[a[i]])
except:
# failure -> next try
i-=1
j=a.rfind('(')+1
# replace brackets in parallel (this part is executed first)
if j:
k=a.find(')',j)
a=a[:j-1]+str(g(a[j:k]))+a[k+1:]
return float(a.replace('--',''))
``````

FWIW, the n+1th Python solution. In a blatant abuse of try-except I use a trial-and-error approach. It should handle all cases properly including stuff like `-(8)`, `--8` and `g('-(1 - 3)')`. It is re-entrant. Without support for the `--` case which many implementations don't support, it is at 217 chars (see previous revision).

Thanks for an interesting hour on a Sunday and another 30 mins on Monday. Thanks to krubo for his nice dict.

Another interesting approach... Identical in length to one of the other Python solutions, too. This confirms my view that using a dictionary of operators is the way to go, where possible. I wanted to do something similar in C#, but the syntax simply takes up too many chars.
+5  A:

# Ruby

Number of characters: 170

Obfuscated:

``````def s(x)
while x.sub!(/\(([^\(\)]*?)\)/){s(\$1)}
x.gsub!('--','')
end
while x.sub!(/(-?[\d.]+)[ ]*([+\-*\/])[ ]*(-?[\d.]+)/){\$1.to_f.send(\$2,\$3.to_f)}
end
x.strip.to_f
end
``````

``````def s(x)
while x.sub!(/\(([^\(\)]*?)\)/){s(\$1)}
x.gsub!('--','')
end
while x.sub!(/(-?[\d.]+)[ ]*([+\-*\/])[ ]*(-?[\d.]+)/){\$1.to_f.send(\$2,\$3.to_f)}
end
x.strip.to_f
end

[
['1 + 3 / -8', -0.5],
['2*3*4*5+99', 219],
['4 * (9 - 4) / (2 * 6 - 2) + 8', 10],
['1 + ((123 * 3 - 69) / 100)', 4],
['2.45/8.5*9.27+(5*0.0023)',2.68344117647059],
['(3+7) - (5+2)', 3]
].each do |pair|
a,b = s(String.new(pair[0])),pair[1]
print pair[0].ljust(25), ' = ', b, ' (', a==b, ')'
puts
end
``````

There is no real obfuscation to this one, which I decided to post fresh since it's wildly different from my first. I should have seen this from the start. The process is a very simple process of elimination: find and resolve the highest pair of parenthesis (the most nested) into a number until no more are found, then resolve all the existing numbers and operations into the result. And, while resolving parenthetical statements I have it strip all double-dashes (Float.to_f doesn't know what to do with them).

So, it supports positive and negative numbers (+3, 3, & -3) and even negated sub-expressions within the parenthesis just by the order of processing. The only shorter implementation is the Perl (w/o eval) one.

Edit: I'm still chasing Perl, but this is the second smallest answer right now. I shrunk it with changes to the second regex and by changing the treatment of the string to be destructive (replaces the old string). This eliminated the need to duplicate the string, which I found out to just be a new pointer to the string. And renaming the function to s from solve saved a few characters.

Nice work, surprised I didn't try this approach myself, since I used something very similar to solve another parsing question.
See my solution for a way to compress that regexp. You shouldn't need the final 'strip' either. And it doesn't look as though you fully implement unary minus, so you get little benefit from the gsub('--','').
I cannot actually shorten my particular algorithm or I fail 3-4 of the tests, I'm not certain why. I could shrink it by maybe 20 characters though.
+5  A:

## Python (without importing anything)

Number of characters: 222

I stole many tricks from Dave's answer, but I managed to shave off some more characters.

``````def e(s,l=0,n=0,f='+'):
if s:l=[c for c in s+')'if' '!=c]
while f!=')':
p=l.pop;m=p(0)
if m=='(':m=e(0,l)
while l[0]not in'+-*/)':m+=p(0)
m=float(m);n={'+':n+m,'-':n-m,'*':n*m,'/':n/(m or 1)}[f];f=p(0)
return n
``````

Commented version:

``````def evaluate(stringexpr, listexpr=0, n=0, f_operation='+'):
# start out as taking 0 + the expression... (or could use 1 * ;)

# We'll prefer to keep the expression as a list of characters,
# so we can use .pop(0) to eat up the expression as we go.
if stringexpr:
listexpr = [c for c in stringexpr+')' if c!=' ']

# use ')' as sentinel to return the answer
while f_operation != ')':
m_next = listexpr.pop(0)
if m_next == '(':
# lists are passed by reference, so this call will eat the (parexp)
m_next = evaluate(None, listexpr)

else:
# rebuild any upcoming numeric chars into a string
while listexpr[0] not in '+-*/)':
m_next += listexpr.pop(0)

# Update n as the current answer.  But never divide by 0.
m = float(m_next)
n = {'+':n+m, '-':n-m, '*':n*m, '/':n/(m or 1)}[f_operation]

# prepare the next operation (known to be one of '+-*/)')
f_operation = listexpr.pop(0)

return n
``````
+1 Nice dict idea. The current version fails on e('1+0'), however. Use {'+':n+m,'-':n-m,'*':n*m,'/':n/m if m else 1} instead. I have borrowed your idea (with this amendment). Thanks
Thanks. I hadn't thought of the DivZero problem; a 7-character fix is n/(m or 1).
Will do this for my program, too ;-)
hehe, don't change anything now, the number of characters is beautiful :)
+6  A:

## J

### Number of characters: 208

After Jeff Moser's comment, I realized that I had completely forgotten about this language... I'm no expert, but my first attempt went rather well.

``````e=:>@{:@[email protected];:
f=:''&(4 :0)
'y x'=.x g y
while.(\$y)*-.')'={.>{.y do.'y x'=.(x,>(-.'/'={.>{.y){('%';y))g}.y end.y;x
)
g=:4 :0
z=.>{.y
if.z='('do.'y z'=.f}.y else.if.z='-'do.z=.'_',>{.}.y end.end.(}.y);":".x,z
)
``````

It's a bit annoying, having to map `x/y` and `-z` into J's `x%y` and `_z`. Without that, maybe 50% of this code could disappear.

Yeah, that's pretty nice. Now what about a solution in K? :P I'm suspecting that might even be able to beat Perl.
Woohoo, I managed to get my Haskell solution under my J solution! Though if somebody here were a J or K or APL wizard, they'd probably destroy the 200-character barrier...
+24  A:

## Visual Basic.NET

Number of characters: 9759

I'm more of a bowler myself.

NOTE: does not take nested parentheses into account. Also, untested, but I'm pretty sure it works.

``````Imports Microsoft.VisualBasic
Imports System.Text
Imports System.Collections.Generic
Public Class Main
Public Shared Function DoArithmaticFunctionFromStringInput(ByVal MathematicalString As String) As Double
Dim numberList As New List(Of Number)
Dim operationsList As New List(Of IOperatable)
Dim currentNumber As New Number
Dim currentParentheticalStatement As New Parenthetical
Dim isInParentheticalMode As Boolean = False
Dim allCharactersInString() As Char = MathematicalString.ToCharArray
For Each mathChar In allCharactersInString
If mathChar = Number.ZERO_STRING_REPRESENTATION Then
currentNumber.UpdateNumber(mathChar)
ElseIf mathChar = Number.ONE_STRING_REPRESENTATION Then
currentNumber.UpdateNumber(mathChar)
ElseIf mathChar = Number.TWO_STRING_REPRESENTATION Then
currentNumber.UpdateNumber(mathChar)
ElseIf mathChar = Number.THREE_STRING_REPRESENTATION Then
currentNumber.UpdateNumber(mathChar)
ElseIf mathChar = Number.FOUR_STRING_REPRESENTATION Then
currentNumber.UpdateNumber(mathChar)
ElseIf mathChar = Number.FIVE_STRING_REPRESENTATION Then
currentNumber.UpdateNumber(mathChar)
ElseIf mathChar = Number.SIX_STRING_REPRESENTATION Then
currentNumber.UpdateNumber(mathChar)
ElseIf mathChar = Number.SEVEN_STRING_REPRESENTATION Then
currentNumber.UpdateNumber(mathChar)
ElseIf mathChar = Number.EIGHT_STRING_REPRESENTATION Then
currentNumber.UpdateNumber(mathChar)
ElseIf mathChar = Number.NINE_STRING_REPRESENTATION Then
currentNumber.UpdateNumber(mathChar)
ElseIf mathChar = Number.DECIMAL_POINT_STRING_REPRESENTATION Then
currentNumber.UpdateNumber(mathChar)

If Not isInParentheticalMode Then
Else
End If

currentNumber = New Number
ElseIf mathChar = Number.NEGATIVE_NUMBER_STRING_REPRESENTATION Then
If currentNumber.StringOfNumbers.Length > 0 Then
currentNumber.UpdateNumber(mathChar)

If Not isInParentheticalMode Then
Else
End If

currentNumber = New Number
Else
currentNumber.UpdateNumber(mathChar)
End If
ElseIf mathChar = Multiplication.MULTIPLICATION_STRING_REPRESENTATION Then
Dim multiplication As New Multiplication

If Not isInParentheticalMode Then
Else
End If
currentNumber = New Number
ElseIf mathChar = Division.DIVISION_STRING_REPRESENTATION Then
Dim division As New Division

If Not isInParentheticalMode Then
Else
End If
currentNumber = New Number
ElseIf mathChar = Parenthetical.LEFT_PARENTHESIS_STRING_REPRESENTATION Then
isInParentheticalMode = True
ElseIf mathChar = Parenthetical.RIGHT_PARENTHESIS_STRING_REPRESENTATION Then
currentNumber = currentParentheticalStatement.EvaluateParentheticalStatement
isInParentheticalMode = False
End If
Next

Dim result As Double = 0
Dim operationIndex As Integer = 0
For Each numberOnWhichToPerformOperations As Number In numberList
result = operationsList(operationIndex).PerformOperation(result, numberOnWhichToPerformOperations)
operationIndex = operationIndex + 1
Next

Return result

End Function
Public Class Number
Public Const DECIMAL_POINT_STRING_REPRESENTATION As Char = "."
Public Const NEGATIVE_NUMBER_STRING_REPRESENTATION As Char = "-"
Public Const ZERO_STRING_REPRESENTATION As Char = "0"
Public Const ONE_STRING_REPRESENTATION As Char = "1"
Public Const TWO_STRING_REPRESENTATION As Char = "2"
Public Const THREE_STRING_REPRESENTATION As Char = "3"
Public Const FOUR_STRING_REPRESENTATION As Char = "4"
Public Const FIVE_STRING_REPRESENTATION As Char = "5"
Public Const SIX_STRING_REPRESENTATION As Char = "6"
Public Const SEVEN_STRING_REPRESENTATION As Char = "7"
Public Const EIGHT_STRING_REPRESENTATION As Char = "8"
Public Const NINE_STRING_REPRESENTATION As Char = "9"

Private _isNegative As Boolean
Public ReadOnly Property IsNegative() As Boolean
Get
Return _isNegative
End Get
End Property
Public ReadOnly Property ActualNumber() As Double
Get
Dim result As String = ""
If HasDecimal Then
If DecimalIndex = StringOfNumbers.Length - 1 Then
result = StringOfNumbers.ToString
Else
result = StringOfNumbers.Insert(DecimalIndex, DECIMAL_POINT_STRING_REPRESENTATION).ToString
End If
Else
result = StringOfNumbers.ToString
End If
If IsNegative Then
result = NEGATIVE_NUMBER_STRING_REPRESENTATION & result
End If
Return CType(result, Double)
End Get
End Property
Private _hasDecimal As Boolean
Public ReadOnly Property HasDecimal() As Boolean
Get
Return _hasDecimal
End Get
End Property
Private _decimalIndex As Integer
Public ReadOnly Property DecimalIndex() As Integer
Get
Return _decimalIndex
End Get
End Property
Private _stringOfNumbers As New StringBuilder
Public ReadOnly Property StringOfNumbers() As StringBuilder
Get
Return _stringOfNumbers
End Get
End Property
Public Sub UpdateNumber(ByVal theDigitToAppend As Char)
If IsNumeric(theDigitToAppend) Then
Me._stringOfNumbers.Append(theDigitToAppend)
ElseIf theDigitToAppend = DECIMAL_POINT_STRING_REPRESENTATION Then
Me._hasDecimal = True
Me._decimalIndex = Me._stringOfNumbers.Length
ElseIf theDigitToAppend = NEGATIVE_NUMBER_STRING_REPRESENTATION Then
Me._isNegative = Not Me._isNegative
End If
End Sub
Public Shared Function ConvertDoubleToNumber(ByVal numberThatIsADouble As Double) As Number
Dim numberResult As New Number
For Each character As Char In numberThatIsADouble.ToString.ToCharArray
numberResult.UpdateNumber(character)
Next
Return numberResult
End Function
End Class
Public MustInherit Class Operation
Protected _firstnumber As New Number
Protected _secondnumber As New Number
Public Property FirstNumber() As Number
Get
Return _firstnumber
End Get
Set(ByVal value As Number)
_firstnumber = value
End Set
End Property
Public Property SecondNumber() As Number
Get
Return _secondnumber
End Get
Set(ByVal value As Number)
_secondnumber = value
End Set
End Property
End Class
Public Interface IOperatable
Function PerformOperation(ByVal number1 As Double, ByVal number2 As Number) As Double
End Interface
Inherits Operation
Implements IOperatable
Public Const ADDITION_STRING_REPRESENTATION As String = "+"
Public Sub New()

End Sub
Public Function PerformOperation(ByVal number1 As Double, ByVal number2 As Number) As Double Implements IOperatable.PerformOperation
Dim result As Double = 0
result = number1 + number2.ActualNumber
Return result
End Function
End Class
Public Class Multiplication
Inherits Operation
Implements IOperatable
Public Const MULTIPLICATION_STRING_REPRESENTATION As String = "*"
Public Sub New()

End Sub
Public Function PerformOperation(ByVal number1 As Double, ByVal number2 As Number) As Double Implements IOperatable.PerformOperation
Dim result As Double = 0
result = number1 * number2.ActualNumber
Return result
End Function
End Class
Public Class Division
Inherits Operation
Implements IOperatable
Public Const DIVISION_STRING_REPRESENTATION As String = "/"
Public Const DIVIDE_BY_ZERO_ERROR_MESSAGE As String = "I took a lot of time to write this program. Please don't be a child and try to defile it by dividing by zero. Nobody thinks you are funny."
Public Sub New()

End Sub
Public Function PerformOperation(ByVal number1 As Double, ByVal number2 As Number) As Double Implements IOperatable.PerformOperation
If Not number2.ActualNumber = 0 Then
Dim result As Double = 0
result = number1 / number2.ActualNumber
Return result
Else
Dim divideByZeroException As New Exception(DIVIDE_BY_ZERO_ERROR_MESSAGE)
Throw divideByZeroException
End If
End Function
End Class
Public Class Parenthetical
Public Const LEFT_PARENTHESIS_STRING_REPRESENTATION As String = "("
Public Const RIGHT_PARENTHESIS_STRING_REPRESENTATION As String = ")"
Private _allNumbers As New List(Of Number)
Public Property AllNumbers() As List(Of Number)
Get
Return _allNumbers
End Get
Set(ByVal value As List(Of Number))
_allNumbers = value
End Set
End Property
Private _allOperators As New List(Of IOperatable)
Public Property AllOperators() As List(Of IOperatable)
Get
Return _allOperators
End Get
Set(ByVal value As List(Of IOperatable))
_allOperators = value
End Set
End Property
Public Sub New()

End Sub
Public Function EvaluateParentheticalStatement() As Number
Dim result As Double = 0
Dim operationIndex As Integer = 0
For Each numberOnWhichToPerformOperations As Number In AllNumbers
result = AllOperators(operationIndex).PerformOperation(result, numberOnWhichToPerformOperations)
operationIndex = operationIndex + 1
Next

Dim numberToReturn As New Number
numberToReturn = Number.ConvertDoubleToNumber(result)
Return numberToReturn
End Function
End Class
End Class
``````
I also probably could have hit 10k characters if it weren't so late at night :)
Do you know less characters is better? This way they never think vb.net is any good.
@ikke - it was supposed to be as FEW characters as possible? oh dear... someone seems to have missed the point
LOL - this reminds me of a few VB programs I've worked with
My eyes! The goggles do nothing!
ZERO_STRING_REPRESENTATION looks like something that belongs on thedailywtf
@lkke - I think that is the point. See comment from finnw.
+1 this made me laugh more than any other answer on SO. **ever**.
+5  A:

Here comes another one:

## Shell script (using sed+awk)

Number of characters: 295

obfuscated:

``````e(){ a="\$1";while echo "\$a"|grep -q \(;do eval "`echo "\$a"|sed 's/\(.*\)(\([^()]*\))\(.*\)/a="\1\`e \"\2\"\`\3"/'`";done; echo "\$a"|sed 's/\([-+*/]\) *\(-\?\) */ \1 \2/g'|awk '{t=\$1;for(i=2;i<NF;i+=2){j=\$(i+1);if(\$i=="+") t+=j; else if(\$i=="-") t-=j; else if(\$i=="*") t*=j; else t/=j}print t}';}
``````

``````e () {
a="\$1"
# Recursively process bracket-expressions
while echo "\$a"|grep -q \(; do
eval "`echo "\$a"|
sed 's/\(.*\)(\([^()]*\))\(.*\)/a="\1\`e \"\2\"\`\3"/'`"
done
# Compute expression without brackets
echo "\$a"|
sed 's/\([-+*/]\) *\(-\?\) */ \1 \2/g'|
awk '{
t=\$1;
for(i=2;i<NF;i+=2){
j=\$(i+1);
if(\$i=="+") t+=j;
else if(\$i=="-") t-=j;
else if(\$i=="*") t*=j;
else t/=j
}
print t
}'
}
``````

Test:

``````str='  2.45 / 8.5  *  9.27   +    (   5   *  0.0023  ) '
echo "\$str"|bc -l
e "\$str"
``````

Result:

``````2.68344117647058823526
2.68344
``````
I have (almost) no idea how this works, but I am amazed how well a shell script does at this task! Well done indeed.
Well, just remember that many many operating systems use that language/tool mix for many many different tasks :)
+1  A:

# Java

Number of Characters: 376

Updated version, now with more ? operator abuse!

Fully obfuscated solution:

``````static double e(String t){t="("+t+")";for(String s:new String[]{"+","-","*","/","(",")"})t=t.replace(s," "+s+" ");return f(new Scanner(t));}static double f(Scanner s){s.next();double a,v=s.hasNextDouble()?s.nextDouble():f(s);while(s.hasNext("[^)]")){char o=s.next().charAt(0);a=s.hasNextDouble()?s.nextDouble():f(s);v=o=='+'?v+a:o=='-'?v-a:o=='*'?v*a:v/a;}s.next();return v;}
``````

Clear/semi-obfuscated function:

``````static double evaluate(String text) {
text = "(" + text + ")";
for (String s : new String[] {"+", "-", "*", "/", "(", ")" }) {
text = text.replace(s, " " + s + " ");
}
return innerEval(new Scanner(text));
}

static double innerEval(Scanner s) {
s.next();
double arg, val = s.hasNextDouble() ? s.nextDouble() : innerEval(s);
while (s.hasNext("[^)]")) {
char op = s.next().charAt(0);
arg = s.hasNextDouble() ? s.nextDouble() : innerEval(s);
val =
op == '+' ? val + arg :
op == '-' ? val - arg :
op == '*' ? val * arg :
val / arg;
}
s.next();
return val;
}
``````
A:

I'm surprised nobody did it in Lex / Yacc or equivalent.

That would seem to yield the shortest source code that was also readable / maintainable.

I was about to do it with ANTLR, but it was larger that you could think in first instance xD
Lex/Yacc is good for writing real parsers, but if your requirements are simple enough, you can do it better (smaller) without.
@Chris: I'm inclined to agree, but I thought somebody might have tried it.
I might try it, but I've currently got a plain C solution down to 209 characters. Getting there.
+4  A:

## Ruby

Number of characters: 217 179

This is the shortest ruby solution up to now (one heavily based on RegExp yields incorrect answers when string contains few groups of parenthesis) -- no longer true. Solutions based on regex and substitution are shorter. This one is based on stack of accumulators and parses whole expression from left to right. It is re-entrant, and does not modify input string. It could be accused of breaking the rules of not using `eval`, as it calls `Float`'s methods with identical names as their mathematical mnemonics (+,-,/,*).

Obfuscated code (old version, tweaked below):

``````def f(p);a,o=[0],['+']
p.sub(/-/,'+-').scan(/(?:(-?\d+(?:\.\d+)?)|(.))\s*/).each{|n|
q,w=n;case w;when'(';a<<0;o<<'+';when')';q=a.pop;else;o<<w
end if q.nil?;a[-1]=a[-1].method(o.pop).call(q.to_f) if !q.nil?};a[0];end
``````

More obfuscated code:

``````def f(p);a,o=[0],[:+]
p.scan(/(?:(-?\d+(?:\.\d+)?)|(.))\s*/).each{|n|q,w=n;case w
when'(';a<<0;o<<:+;when')';q=a.pop;else;o<<w;end if !q
a<<a.pop.send(o.pop,q.to_f)if q};a[0];end
``````

Clean code:

``````def f(p)
accumulators, operands = [0], ['+']
p.gsub(/-/,'+-').scan(/(?:(-?\d+(?:\.\d+)?)|(.))\s*/).each do |n|
number, operand = n
case operand
when '('
accumulators << 0
operands << '+'
when ')'
number = accumulators.pop
operands.pop
else
operands[-1] = operand
end if number.nil?
accumulators[-1] = accumulators.last.method(operands[-1]).call(number.to_f) unless number.nil?
end
accumulators.first
end
``````
Actually, mine is shorter (198) and uses regex to solve parenthesis from the top down before the final result of the mathematics. So "3 + (3 * (3 + 9))" goes: "3 + (3 * 12)", "3 + 36", 39. It goes top-to-bottom, left-to-right. It solves all tests, except one minor oversight which requires spaces between tokens. See: http://stackoverflow.com/questions/928563/code-golf-evaluating-mathematical-expressions/932627#932627
Not that yours isn't clever, it very much is.
(3+7) - (5+2) -- that's what I've meant by several parentheses groups. Your solution has problem with non-nested parentheses, because of regex greediness.
That may be, but I had been fiddling around with my parser yesterday night and improved it on my system (with mathematical functions and single letter variables). So I pulled my better regex from it and it works just fine, the post is updated with a new char count too. ;-) I'll roll in your equation into the tests in the answer momentarily.
I don't think the use of 'method' or 'send' is cheating - it's just a table lookup - you are not using the built-in parser.
+3  A:

## MATLAB (v7.8.0)

Number of characters: 239

Obfuscated function:

``````function [v,s]=m(s),r=1;while s,s=regexp(s,'( ?)(?(1)-?)[\.\d]+|\S','match');c=s{end};s=[s{1:end-1}];if any(c>47),v=str2num(c);elseif c>41,[l,s]=m(s);v=[l/v l*v l+v l-v];v=v(c=='/*+-');if r,break;end;r=1;elseif c<41,break;end;r=r&c~=41;end
``````

Clear(er) function:

``````function [value,str] = math(str)
returnNow = 1;
while str,
str = regexp(str,'( ?)(?(1)-?)[\.\d]+|\S','match');
current = str{end};
str = [str{1:end-1}];
if any(current > 47),
value = str2num(current);
elseif current > 41,
[leftValue,str] = math(str);
value = [leftValue/value leftValue*value ...
leftValue+value leftValue-value];
value = value(current == '/*+-');
if returnNow,
break;
end;
returnNow = 1;
elseif current < 41,
break;
end;
returnNow = returnNow & (c ~= 41);
end
``````

Test:

``````>> [math('1 + 3 / -8'); ...
math('2*3*4*5+99'); ...
math('4 * (9 - 4) / (2 * 6 - 2) + 8'); ...
math('1 + ((123 * 3 - 69) / 100)'); ...
math('2.45/8.5*9.27+(5*0.0023)')]

ans =

-0.5000
219.0000
10.0000
4.0000
2.6834
``````

Synopsis: A mixture of regular expressions and recursion. Pretty much the best I have been able to do so far, without cheating and using EVAL.

+32  A:

## Assembler

427 bytes

Obfuscated, assembled with the excellent A86 into a .com executable:

``````dd 0db9b1f89h, 081bee3h, 0e8af789h, 0d9080080h, 0bdac7674h, 013b40286h
dd 07400463ah, 0ccfe4508h, 08ce9f675h, 02fc8000h, 013b0057eh, 0feaac42ah
dd 0bedf75c9h, 0ba680081h, 04de801h, 04874f73bh, 04474103ch, 0e8e8b60fh
dd 08e8a003fh, 0e880290h, 0de0153h, 08b57e6ebh, 0d902a93eh, 046d891dh
dd 08906c783h, 05f02a93eh, 03cffcee8h, 057197510h, 02a93e8bh, 08b06ef83h
dd 05d9046dh, 02a93e89h, 03bc9d95fh, 0ac0174f7h, 074f73bc3h, 0f3cac24h
dd 0eed9c474h, 0197f0b3ch, 07cc4940fh, 074f73b09h, 0103cac09h, 0a3ce274h
dd 0e40a537eh, 0e0d90274h, 02a3bac3h, 021cd09b4h, 03e8b20cdh, 0ff8102a9h
dd 0ed7502abh, 0474103ch, 0e57d0b3ch, 0be02a3bfh, 014d903a3h, 0800344f6h
dd 02db00574h, 0d9e0d9aah, 0d9029f2eh, 0bb34dfc0h, 08a0009h, 01c75f0a8h
dd 020750fa8h, 0b0f3794bh, 021e9aa30h, 0de607400h, 08802990eh, 0de07df07h
dd 0c392ebc1h, 0e8c0008ah, 0aa300404h, 0f24008ah, 04baa3004h, 02eb0ee79h
dd 03005c6aah, 0c0d90ab1h, 0e9defcd9h, 02a116deh, 0e480e0dfh, 040fc8045h
dd 0ede1274h, 0c0d90299h, 015dffcd9h, 047300580h, 0de75c9feh, 0303d804fh
dd 03d80fa74h, 04f01752eh, 0240145c6h, 0dfff52e9h, 0d9029906h, 0f73b025fh
dd 03caca174h, 07fed740ah, 0df07889ah, 0277d807h, 047d9c1deh, 0990ede02h
dd 025fd902h, 03130e0ebh, 035343332h, 039383736h, 02f2b2d2eh, 02029282ah
dd 0e9000a09h, 07fc9f9c1h, 04500000fh, 0726f7272h
db 024h, 0abh, 02h
``````

EDIT: Unobfuscated source:

``````  mov [bx],bx
finit
mov si,81h
mov di,si
mov cl,[80h]
or cl,bl
jz ret
l1:
lodsb
mov bp,d1
mov ah,19
l2:
cmp al,[bp]
je l3
inc bp
dec ah
jne l2
jmp exit
l3:
cmp ah,2
jle l4
mov al,19
sub al,ah
stosb
l4:
dec cl
jnz l1
mov si,81h
push done

decode:
l5:
call l7
l50:
cmp si,di
je ret
cmp al,16
je ret
db 0fh, 0b6h, 0e8h ; movzx bp,al
call l7
mov cl,[bp+op-11]
mov byte ptr [sm1],cl
db 0deh
sm1:db ?
jmp l50

open:
push di
mov di,word ptr [s]
fstp dword ptr [di]
mov [di+4],bp
mov word ptr [s],di
pop di
call decode
cmp al,16
jne ret
push di
mov di,word ptr [s]
sub di,6
mov bp,[di+4]
fld dword ptr [di]
mov word ptr [s],di
pop di
fxch st(1)
cmp si,di
je ret
lodsb
ret

l7: cmp si,di
je exit
lodsb
cmp al,15
je open
fldz
cmp al,11
jg exit
db 0fh, 94h, 0c4h ; sete ah
jl l10
l9:
cmp si,di
je l12
lodsb
cmp al,16
je ret
l10:
cmp al,10
jle l12i

l12:
or ah,ah
je l13
fchs
l13:
ret

exit:
mov dx,offset res
mov ah,9
int 21h
int 20h

done:
mov di,word ptr [s]
cmp di,(offset s)+2
jne exit
cmp al,16
je ok
cmp al,11
jge exit
ok:
mov di,res
mov si,res+100h
fst dword ptr [si]
test byte ptr [si+3],80h
jz pos
mov al,'-'
stosb
fchs
pos:
fldcw word ptr [cw]
fld st(0)
fbstp [si]
mov bx,9
l1000:
mov al,[si+bx]
test al,0f0h
jne startu
test al,0fh
jne startl
dec bx
jns l1000
mov al,'0'
stosb
jmp frac

l12i:
je l11
fimul word ptr [d3]
mov [bx],al
fild word ptr [bx]
jmp l9
ret

startu:
mov al,[si+bx]
shr al,4
stosb
startl:
mov al,[si+bx]
and al,0fh
stosb
dec bx
jns startu

frac:
mov al,'.'
stosb
mov byte ptr [di],'0'
mov cl,10
fld st(0)
frndint
frac1:
fsubp st(1)
ficom word ptr [zero]
fstsw ax
and ah,045h
cmp ah,040h
je finished
fimul word ptr [d3]
fld st(0)
frndint
fist word ptr [di]
inc di
dec cl
jnz frac1

finished:
dec di
cmp byte ptr [di],'0'
je finished
cmp byte ptr [di],'.'
jne f2
dec di
f2:
mov byte ptr [di+1],'\$'
exit2:
jmp exit

l11:
fild word ptr [d3]
fstp dword ptr [bx+2]
l111:
cmp si,di
je ret
lodsb
cmp al,10
je exit2
jg ret
mov [bx],al
fild word ptr [bx]
fdiv dword ptr [bx+2]
fld dword ptr [bx+2]
fimul word ptr [d3]
fstp dword ptr [bx+2]
jmp l111

d1: db '0123456789.-+/*()', 32, 9
d3: dw 10
op: db 0e9h, 0c1h, 0f9h, 0c9h
cw: dw 0f7fh
zero: dw 0
res:db 'Error\$'
s:  dw (offset s)+2
``````

Skizz

. O.o .
+1 for being hardcore.
Assembly - this is *real* programming!
I once saw a full tetris game in 64 bytes
+3  A:
+26  A:

## Ruby

Number of characters: 103

``````N='( *-?[\d.]+ *)'
def e x
x.sub!(/\(#{N}\)|#{N}([^.\d])#{N}/){\$1or(e\$2).send\$3,e(\$4)}?e(x):x.to_f
end
``````

This is a non-recursive version of The Wicked Flea's solution. Parenthesized sub-expressions are evaluated bottom-up instead of top-down.

Edit: Converting the 'while' to a conditional + tail recursion has saved a few characters, so it is no longer non-recursive (though the recursion is not semantically necessary.)

Edit: Borrowing Daniel Martin's idea of merging the regexps saves another 11 characters!

Edit: That recursion is even more useful than I first thought! `x.to_f` can be rewritten as `e(x)`, if `x` happens to contain a single number.

Edit: Using '`or`' instead of '`||`' allows a pair of parentheses to be dropped.

Long version:

``````# Decimal number, as a capturing group, for substitution
# in the main regexp below.
N='( *-?[\d.]+ *)'

# The evaluation function
def e(x)
matched = x.sub!(/\(#{N}\)|#{N}([^\d.])#{N}/) do
# Group 1 is a numeric literal in parentheses.  If this is present then
# just return it.
if \$1
\$1
# Otherwise, \$3 is an operator symbol and \$2 and \$4 are the operands
else
# Recursively call e to parse the operands (we already know from the
# regexp that they are numeric literals, and this is slightly shorter
# than using :to_f)
e(\$2).send(\$3, e(\$4))
# We could have converted \$3 to a symbol (\$3.to_s) or converted the
# result back to string form, but both are done automatically anyway
end
end
if matched then
# We did one reduction. Now recurse back and look for more.
e(x)
else
# If the string doesn't look like a non-trivial expression, assume it is a
# string representation of a real number and attempt to parse it
x.to_f
end
end
``````
I almost thought this was the new leader until I saw the Perl one had been edited to become even shorter! Good job, anyway.
Getting rid of 'e=readline.chomp;...;p e.to_f' and using 'def q(e);...;e.to_f;end' like the other solutions would save 10 characters. However, it fails to q("1 + 3 / -8")==-0.5 as in the question.
@ephemient that's a bug you found - it could not handle negative numbers.
The gsub!('--','') in my code is for how the parenthetical argument works, if negated. If the result of the interior of a negated parenthetical is negative the minus outside the statement remains: --7.0, for example. However, supporting that costs me 24 characters, still 19 above you. I don't know that I can shrink it any more than the tricks I learned from you. (But I did great for a 2nd try!)
Using "send" is really coming close to violating the "no eval" rule.But nice trick incorporating the spaces into your number regex. Using that trick and another one got my perl solution down to 119 characters.
Perl's down to 107 now, but that's still 4 + 103...
+1  A:

## C++

Chars: 1670

`````` // not trying to be terse here
#define DIGIT(c)((c)>='0' && (c) <= '9')
#define WHITE(pc) while(*pc == ' ') pc++
#define LP '('
#define RP ')'

bool SeeNum(const char* &pc, float& fNum){
WHITE(pc);
if (!(DIGIT(*pc) || (*pc=='.'&& DIGIT(pc[1])))) return false;
const char* pc0 = pc;
while(DIGIT(*pc)) pc++;
if (*pc == '.'){
pc++;
while(DIGIT(*pc)) pc++;
}
char buf[200];
int len = pc - pc0;
strncpy(buf, pc0, len); buf[len] = 0;
fNum = atof(buf);
return true;
}

bool SeeChar(const char* &pc, char c){
WHITE(pc);
if (*pc != c) return false;
pc++;
return true;
}

void ParsExpr(const char* &pc, float &fNum);

void ParsPrim(const char* &pc, float &fNum){
if (SeeNum(pc, fNum));
else if (SeeChar(pc, LP)){
ParsExpr(pc, fNum);
if (!SeeChar(pc, RP)) exit(0);
}
else exit(0); // you can abort better than this
}

void ParsUnary(const char* &pc, float &fNum){
if (SeeChar(pc, '-')){
pc+;
ParsUnary(pc, fNum);
fNum = -fNum;
}
else {
ParsPrim(pc, fNum);
}
}

void ParsExpr(const char* &pc, float &fNum){
ParsUnary(pc, fNum);
float f1 = 0;
while(true){
if (SeeChar(pc, '+')){
ParsUnary(pc, f1);
fNum += f1;
}
else if (SeeChar(pc, '-')){
ParsUnary(pc, f1);
fNum -= f1;
}
else if (SeeChar(pc, '*')){
ParsUnary(pc, f1);
fNum *= f1;
}
else if (SeeChar(pc, '/')){
ParsUnary(pc, f1);
fNum /= f1;
}
else break;
}
}
``````

This is just LL1 (recursive descent). I like to do it this way (although I use doubles) because it's plenty fast, and easy to insert routines to handle precedence levels.

+3  A:

C#

Number of Characters: 355

I took Noldorin's Answer and modified it, so give Noldorin 99% of the credit for this. Best I could do with the algorithm was using was 408 characters. See Noldorin's Answer for the clearer code version.

Change char comparisons to compare against numbers.
Removed some default declarations and combined same type of declarations.
Re-worked some of the if statments.

``````float q(string x){float v,n;if(!float.TryParse(x,out v)){x+=';';int t=0,l=0,i=0;char o,s='?',p='+';for(;i<x.Length;i++){o=s;if(x[i]!=32){s=x[i];if(char.IsDigit(x[i])|s==46|(s==45&o!=49))s='1';if(s==41)l--;if(s!=o&l==0){if(o==49|o==41){n=q(x.Substring(t,i-t));v=p==43?v+n:p==45?v-n:p==42?v*n:p==47?v/n:v;p=x[i];}t=i;if(s==40)t++;}if(s==40)l++;}}}return v;}
``````

Edit: knocked it down some more, from 361 to 355, by removing one of the return statments.

Ah, I didn't realise you'd already posted it as a new answer. Thanks for all the credit (which is probably more than I deserve, as I was stuck around 390). I'll take a look more closely at the modifications soon... the only one that I considered was changing char comparisons to use numbers. :)
+2  A:

## C# with Regex

Number of characters: 294

This is partially based off Jeff Moser's answer, but with a significantly simplified evaluation technique. There might even be further ways to reduce the char count, but I'm quite pleased now that there's a C# solution under 300 chars!

Fully obfuscated code:

``````float e(string x){while(x.Contains("("))x=Regex.Replace(x,@"\(([^\(]*?)\)",m=>e(m.Groups[1].Value).ToString());float r=0;foreach(Match m in Regex.Matches("+"+x,@"\D ?-?[\d.]+")){var o=m.Value[0];var v=float.Parse(m.Value.Substring(1));r=o=='+'?r+v:o=='-'?r-v:o=='*'?r*v:r/v;}return r;}
``````

Clearer code:

``````float e(string x)
{
while (x.Contains("("))
x = Regex.Replace(x, @"\(([^\(]*?)\)", m => e(m.Groups[1].Value).ToString());
float r = 0;
foreach (Match m in Regex.Matches("+" + x, @"\D ?-?[\d.]+"))
{
var o = m.Value[0];
var v = float.Parse(m.Value.Substring(1));
r = o == '+' ? r + v : o == '-' ? r - v : o == '*' ? r * v : r / v;
}
return r;
}
``````
+21  A:

# C (VS2005)

Number of Characters: 1360

Abuse of preprocessor and warnings for fun code layout (scroll down to see):

``````#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define b main
#define c(a) b(a,0)
#define d -1
#define e -2
#define g break
#define h case
#define hh h
#define hhh h
#define w(i) case i
#define i return
#define j switch
#define k float
#define l realloc
#define m sscanf
#define n int _
#define o char
#define t(u) #u
#define q(r) "%f" t(r)  "n"
#define s while
#define v default
#define ex exit
#define W printf
#define x fn()
#define y strcat
#define z strcpy
#define Z strlen

char*p    =0    ;k    *b    (n,o**    a){k*f
;j(_){    hh   e:     i*    p==40?    (++p,c
(d        ))  :(      f=        l(        0,
4)        ,m (p       ,q        (%        ),
f,&_),    p+=_        ,f       );        hh
d:f=c(    e);s        (1      ){        j(
*p    ++ ){       hh     0:        hh
41    :i  f;      hh    43        :*
f+=*c(    e)   ;g     ;h    45:*f=    *f-*c(
e);g;h    42    :*    f=    *f**c(    e);g;h

47:*f      /=*c      (e);     g;   v:    c(0);}
}w(1):    if(p&&    printf    (q  ((     "\\"))
,*  c(    d)  ))    g;  hh    0: ex      (W
(x  ))    ;v  :p    =(        p?y:       z)(l(p
,Z(1[     a]  )+    (p        ?Z(p           )+
1:1))     ,1  [a    ])  ;b    (_ -1          ,a
+1  );    g;  }i    0;};fn    ()  {n     =42,p=
43  ;i     "Er"      "ro"     t(   r)    "\n";}
``````

Skizz

And how long did that take you? =)
Far too long ;-)
This is soooo cute!!
+1  A:

## PowerBASIC

Number of characters: ~400

A bit ugly, but it works. :) I'm sure regexp would have made it even smaller.

``````DEFDBL E,f,i,z,q,a,v,o
DEFSTR s,c,k,p

FUNCTION E(s)

i=LEN(s)
DO
IF MID\$(s,i,1)="("THEN
q=INSTR(i,s,")")
s=LEFT\$(s,i-1)+STR\$(E(MID\$(s,i+1,q-i-1)))+MID\$(s,q+1)
END IF
i-=1
LOOP UNTIL i=0

k="+-*/"
DIM p(PARSECOUNT(s,ANY k))
PARSE s,p(),ANY k

a=VAL(p(0))

FOR i=1TO LEN(s)
c=MID\$(s,i,1)
q=INSTR(k,c)
IF q THEN
z+=1
IF o=0 THEN o=q ELSE p(z)=c+p(z)
IF TRIM\$(p(z))<>"" THEN
v=VAL(p(z))
a=CHOOSE(o,a+v,a-v,a*v,a/v)
o=0
END IF
END IF
NEXT

E=a
END FUNCTION
``````
+1  A:

## C#, 264 characters

Strategy: the first 2 lines get rid of parentheses by induction. Then I split by `\-?[\d.]+` to get numbers and operators. then using aggregate to reduce the string array to a double value.

Variable explanations

m is parenthesized expression with no nested parentheses.
d is a placeholder for that awkward TryParse syntax.
v is the accumulator for the final value
t is the current token.

``````float E(string s){var d=999f;while(d-->1)s=Regex.Replace(s,@"(([^(]?))",m=>E(m.Groups[1].Value)+"");return Regex.Split(s,@"(-?[\d.]+)").Aggregate(d,(v,t)=>(t=t.Trim()).Length==0?v:!float.TryParse(t,out d)?(s=t)==""?0:v:s=="/"?v/d:s=="-"?v-d:s==""?v*d:v+d);}
``````

``````    float F(string s) {
var d=999f;
while(d-->1)
s=Regex.Replace(s,@"\(([^\(]*?)\)",m=>F(m.Groups[1].Value)+"");
return Regex.Split(s, @"(\-?[\d\.]+)")
.Aggregate(d, (v, t) =>
(t=t.Trim()).Length == 0 ? v :
!float.TryParse(t, out d) ? (s=t) == "" ? 0 : v :
s == "/" ? v / d :
s == "-" ? v - d :
s == "*" ? v * d :
v + d);
}
``````

EDIT: shamelessly stole parts from noldorin's answer, reused s as the operator variable.

EDIT: 999 nested parentheses should be enough for anyone.

Pretty nice. Looks fairly similar to my regex solution in fact. It's all shown me a few interesting tricks for reducing count such as `+""` instead of `ToString`. Well done for throwing in the LINQ, too.
wow, I didn't see that one :) I might not have bothered posting this answer if I had. I think a lot of the differences simply come from my limited understanding of regex
Hehe, no worries. I'm just quite surprised how similar many parts of our solution are. You've certainly tempted me to reduce my char count now... I know I certainly can a bit.
+5  A:

## SQL (SQL Server 2008)

### Number of characters: 4202

Fully obfuscated function:

``````WITH Input(id,str)AS(SELECT 1,'1 + 3 / -8'UNION ALL SELECT 2,'2*3*4*5+99'UNION ALL SELECT 3,'4 * (9 - 4)/ (2 * 6 - 2)+ 8'UNION ALL SELECT 4,'1 + ((123 * 3 - 69)/ 100)'UNION ALL SELECT 5,'2.45/8.5*9.27+(5*0.0023)'),Separators(i,ch,str_src,priority)AS(SELECT 1,'-',1,1UNION ALL SELECT 2,'+',1,1UNION ALL SELECT 3,'*',1,1UNION ALL SELECT 4,'/',1,1UNION ALL SELECT 5,'(',0,0UNION ALL SELECT 6,')',0,0),SeparatorsStrSrc(str,i)AS(SELECT CAST('['AS varchar(max)),0UNION ALL SELECT str+ch,SSS.i+1FROM SeparatorsStrSrc SSS INNER JOIN Separators S ON SSS.i=S.i-1WHERE str_src<>0),SeparatorsStr(str)AS(SELECT str+']'FROM SeparatorsStrSrc WHERE i=(SELECT COUNT(*)FROM Separators WHERE str_src<>0)),ExprElementsSrc(id,i,tmp,ele,pre_ch,input_str)AS(SELECT id,1,CAST(LEFT(str,1)AS varchar(max)),CAST(''AS varchar(max)),CAST(' 'AS char(1)),SUBSTRING(str,2,LEN(str))FROM Input UNION ALL SELECT id,CASE ele WHEN''THEN i ELSE i+1 END,CAST(CASE WHEN LEFT(input_str,1)=' 'THEN''WHEN tmp='-'THEN CASE WHEN pre_ch LIKE(SELECT str FROM SeparatorsStr)THEN tmp+LEFT(input_str,1)ELSE LEFT(input_str,1)END WHEN LEFT(input_str,1)IN(SELECT ch FROM Separators)OR tmp IN(SELECT ch FROM Separators)THEN LEFT(input_str,1)ELSE tmp+LEFT(input_str,1)END AS varchar(max)),CAST(CASE WHEN LEFT(input_str,1)=' 'THEN tmp WHEN LEFT(input_str,1)='-'THEN CASE WHEN tmp IN(SELECT ch FROM Separators)THEN tmp ELSE''END WHEN LEFT(input_str,1)IN(SELECT ch FROM Separators)OR tmp IN(SELECT ch FROM Separators)THEN CASE WHEN tmp='-'AND pre_ch LIKE(SELECT str FROM SeparatorsStr)THEN''ELSE tmp END ELSE''END AS varchar(max)),CAST(LEFT(ele,1)AS char(1)),SUBSTRING(input_str,2,LEN(input_str))FROM ExprElementsSrc WHERE input_str<>''OR tmp<>''),ExprElements(id,i,ele)AS(SELECT id,i,ele FROM ExprElementsSrc WHERE ele<>''),Scanner(id,i,val)AS(SELECT id,i,CAST(ele AS varchar(max))FROM ExprElements WHERE ele<>''UNION ALL SELECT id,MAX(i)+1,NULL FROM ExprElements GROUP BY id),Operator(op,priority)AS(SELECT ch,priority FROM Separators WHERE priority<>0),Calc(id,c,i,pop_count,s0,s1,s2,stack,status)AS(SELECT Scanner.id,1,1,0,CAST(scanner.val AS varchar(max)),CAST(NULL AS varchar(max)),CAST(NULL AS varchar(max)),CAST(''AS varchar(max)),CAST('init'AS varchar(max))FROM Scanner WHERE Scanner.i=1UNION ALL SELECT Calc.id,Calc.c+1,Calc.i,3,NULL,NULL,NULL,CASE Calc.s1 WHEN'+'THEN CAST(CAST(Calc.s2 AS real)+CAST(Calc.s0 AS real)AS varchar(max))WHEN'-'THEN CAST(CAST(Calc.s2 AS real)-CAST(Calc.s0 AS real)AS varchar(max))WHEN'*'THEN CAST(CAST(Calc.s2 AS real)*CAST(Calc.s0 AS real)AS varchar(max))WHEN'/'THEN CAST(CAST(Calc.s2 AS real)/CAST(Calc.s0 AS real)AS varchar(max))ELSE NULL END+' '+stack,CAST('calc '+Calc.s1 AS varchar(max))FROM Calc INNER JOIN Scanner NextVal ON Calc.id=NextVal.id AND Calc.i+1=NextVal.i WHERE Calc.pop_count=0AND ISNUMERIC(Calc.s2)=1AND Calc.s1 IN(SELECT op FROM Operator)AND ISNUMERIC(Calc.s0)=1AND(SELECT priority FROM Operator WHERE op=Calc.s1)>=COALESCE((SELECT priority FROM Operator WHERE op=NextVal.val),0)UNION ALL SELECT Calc.id,Calc.c+1,Calc.i,3,NULL,NULL,NULL,s1+' '+stack,CAST('paren'AS varchar(max))FROM Calc WHERE pop_count=0AND s2='('AND ISNUMERIC(s1)=1AND s0=')'UNION ALL SELECT Calc.id,Calc.c+1,Calc.i,Calc.pop_count-1,s1,s2,CASE WHEN LEN(stack)>0THEN SUBSTRING(stack,1,CHARINDEX(' ',stack)-1)ELSE NULL END,CASE WHEN LEN(stack)>0THEN SUBSTRING(stack,CHARINDEX(' ',stack)+1,LEN(stack))ELSE''END,CAST('pop'AS varchar(max))FROM Calc WHERE Calc.pop_count>0UNION ALL SELECT Calc.id,Calc.c+1,Calc.i+1,Calc.pop_count,CAST(NextVal.val AS varchar(max)),s0,s1,coalesce(s2,'')+' '+stack,cast('read'as varchar(max))FROM Calc INNER JOIN Scanner NextVal ON Calc.id=NextVal.id AND Calc.i+1=NextVal.i WHERE NextVal.val IS NOT NULL AND Calc.pop_count=0AND((Calc.s0 IS NULL OR calc.s1 IS NULL OR calc.s2 IS NULL)OR NOT(ISNUMERIC(Calc.s2)=1AND Calc.s1 IN(SELECT op FROM Operator)AND ISNUMERIC(calc.s0)=1AND (SELECT priority FROM Operator WHERE op=Calc.s1)>=COALESCE((SELECT priority FROM Operator WHERE op=NextVal.val),0))AND NOT(s2='('AND ISNUMERIC(s1)=1AND s0=')')))SELECT Calc.id,Input.str,Calc.s0 AS result FROM Calc INNER JOIN Input ON Calc.id=Input.id WHERE Calc.c=(SELECT MAX(c)FROM Calc calc2 WHERE Calc.id=Calc2.id)ORDER BY id
``````

Clear/semi-obfuscated function:

``````WITH
Input(id, str) AS (
SELECT 1, '1 + 3 / -8'
UNION ALL SELECT 2, '2*3*4*5+99'
UNION ALL SELECT 3, '4 * (9 - 4) / (2 * 6 - 2) + 8'
UNION ALL SELECT 4, '1 + ((123 * 3 - 69) / 100)'
UNION ALL SELECT 5, '2.45/8.5*9.27+(5*0.0023)'
)
, Separators(i, ch, str_src, priority) AS (
SELECT 1, '-', 1, 1
UNION ALL SELECT 2, '+', 1, 1
UNION ALL SELECT 3, '*', 1, 1
UNION ALL SELECT 4, '/', 1, 1
UNION ALL SELECT 5, '(', 0, 0
UNION ALL SELECT 6, ')', 0, 0
)
SELECT CAST('[' AS varchar(max)), 0
UNION ALL
SELECT
str + ch
, SSS.i + 1
FROM
INNER JOIN Separators S ON SSS.i = S.i - 1
WHERE
str_src <> 0
)
SELECT str + ']' FROM SeparatorsStrSrc
WHERE i = (SELECT COUNT(*) FROM Separators WHERE str_src <> 0)
)
, ExprElementsSrc(id, i, tmp, ele, pre_ch, input_str) AS (
SELECT
id
, 1
, CAST(LEFT(str, 1) AS varchar(max))
, CAST('' AS varchar(max))
, CAST(' ' AS char(1))
, SUBSTRING(str, 2, LEN(str))
FROM
Input
UNION ALL
SELECT
id
, CASE ele
WHEN '' THEN i
ELSE i + 1
END
, CAST(
CASE
WHEN LEFT(input_str, 1) = ' '
THEN ''
WHEN tmp = '-'
THEN CASE
WHEN pre_ch LIKE (SELECT str FROM SeparatorsStr)
THEN tmp + LEFT(input_str, 1)
ELSE LEFT(input_str, 1)
END
WHEN LEFT(input_str, 1) IN (SELECT ch FROM Separators)
OR
tmp IN (SELECT ch FROM Separators)
THEN LEFT(input_str, 1)
ELSE tmp + LEFT(input_str, 1)
END
AS varchar(max))
, CAST(
CASE
WHEN LEFT(input_str, 1) = ' '
THEN tmp
WHEN LEFT(input_str, 1) = '-'
THEN CASE
WHEN tmp IN (SELECT ch FROM Separators)
THEN tmp
ELSE ''
END
WHEN LEFT(input_str, 1) IN (SELECT ch FROM Separators)
OR
tmp IN (SELECT ch FROM Separators)
THEN CASE
WHEN tmp = '-' AND pre_ch LIKE (SELECT str FROM SeparatorsStr)
THEN ''
ELSE tmp
END
ELSE ''
END
AS varchar(max))
, CAST(LEFT(ele, 1) AS char(1))
, SUBSTRING(input_str, 2, LEN(input_str))
FROM
ExprElementsSrc
WHERE
input_str <> ''
OR
tmp <> ''
)
, ExprElements(id, i, ele) AS (
SELECT
id
, i
, ele
FROM
ExprElementsSrc
WHERE
ele <> ''
)
, Scanner(id, i, val) AS (
SELECT
id
, i
, CAST(ele AS varchar(max))
FROM
ExprElements
WHERE
ele <> ''
UNION ALL
SELECT
id
, MAX(i) + 1
, NULL
FROM
ExprElements
GROUP BY
id
)
, Operator(op, priority) AS (
SELECT
ch
, priority
FROM
Separators
WHERE
priority <> 0
)
, Calc(id, c, i, pop_count, s0, s1, s2, stack, status) AS (
SELECT
Scanner.id
, 1
, 1
, 0
, CAST(scanner.val AS varchar(max))
, CAST(NULL AS varchar(max))
, CAST(NULL AS varchar(max))
, CAST('' AS varchar(max))
, CAST('init' AS varchar(max))
FROM
Scanner
WHERE
Scanner.i = 1
UNION ALL
SELECT
Calc.id
, Calc.c + 1
, Calc.i
, 3
, NULL
, NULL
, NULL
, CASE Calc.s1
WHEN '+' THEN CAST(CAST(Calc.s2 AS real) + CAST(Calc.s0 AS real) AS varchar(max))
WHEN '-' THEN CAST(CAST(Calc.s2 AS real) - CAST(Calc.s0 AS real) AS varchar(max))
WHEN '*' THEN CAST(CAST(Calc.s2 AS real) * CAST(Calc.s0 AS real) AS varchar(max))
WHEN '/' THEN CAST(CAST(Calc.s2 AS real) / CAST(Calc.s0 AS real) AS varchar(max))
ELSE NULL
END
+ ' '
+ stack
, CAST('calc ' + Calc.s1 AS varchar(max))
FROM
Calc
INNER JOIN Scanner NextVal ON Calc.id = NextVal.id
AND Calc.i + 1 = NextVal.i
WHERE
Calc.pop_count = 0
AND ISNUMERIC(Calc.s2) = 1
AND Calc.s1 IN (SELECT op FROM Operator)
AND ISNUMERIC(Calc.s0) = 1
AND (SELECT priority FROM Operator WHERE op = Calc.s1)
>= COALESCE((SELECT priority FROM Operator WHERE op = NextVal.val), 0)
UNION ALL
SELECT
Calc.id
, Calc.c + 1
, Calc.i
, 3
, NULL
, NULL
, NULL
, s1 + ' ' + stack
, CAST('paren' AS varchar(max))
FROM
Calc
WHERE
pop_count = 0
AND s2 = '('
AND ISNUMERIC(s1) = 1
AND s0 = ')'
UNION ALL
SELECT
Calc.id
, Calc.c + 1
, Calc.i
, Calc.pop_count - 1
, s1
, s2
, CASE
WHEN LEN(stack) > 0
THEN SUBSTRING(stack, 1, CHARINDEX(' ', stack) - 1)
ELSE NULL
END
, CASE
WHEN LEN(stack) > 0
THEN SUBSTRING(stack, CHARINDEX(' ', stack) + 1, LEN(stack))
ELSE ''
END
, CAST('pop' AS varchar(max))
FROM
Calc
WHERE
Calc.pop_count > 0
UNION ALL
SELECT
Calc.id
, Calc.c + 1
, Calc.i + 1
, Calc.pop_count
, CAST(NextVal.val AS varchar(max))
, s0
, s1
, coalesce(s2, '') + ' ' + stack
FROM
Calc
INNER JOIN Scanner NextVal ON Calc.id = NextVal.id
AND Calc.i + 1 = NextVal.i
WHERE
NextVal.val IS NOT NULL
AND Calc.pop_count = 0
AND (
(Calc.s0 IS NULL or calc.s1 is null or calc.s2 is null)
OR
NOT(
ISNUMERIC(Calc.s2) = 1
AND Calc.s1 IN (SELECT op FROM Operator)
AND ISNUMERIC(calc.s0) = 1
AND (SELECT priority FROM Operator WHERE op = Calc.s1)
>= COALESCE((SELECT priority FROM Operator WHERE op = NextVal.val), 0)
)
AND NOT(s2 = '(' AND ISNUMERIC(s1) = 1 AND s0 = ')')
)
)
SELECT
Calc.id
, Input.str
, Calc.s0 AS result
FROM
Calc
INNER JOIN Input ON Calc.id = Input.id
WHERE
Calc.c = (SELECT MAX(c) FROM Calc calc2
WHERE Calc.id = Calc2.id)
ORDER BY
id
``````

It is not shortest. But I think that it is very flexible for SQL. It's easy to add new operators. It's easy to change priority of operators.

Gosh, I don't think I was ever expecting an SQL solution! This isn't completely in the spirit of code golf, but up-voted anyway for the audacity (and not even using a programming language). :)
@Noldorin, why isn't it in the spirit of code golf?
A:

## PHP

Number of characters: 170

Fully obfuscated function:

``````function a(\$a,\$c='#\(([^()]*)\)#e',\$d='a("\$1","#^ *-?[\d.]+ *\S *-?[\d.]+ *#e","\\$0")'){\$e='preg_replace';while(\$a!=\$b=\$e(\$c,\$d,\$a))\$a = \$b;return\$e('#^(.*)\$#e',\$d,\$a);}
``````

Clearer function:

``````function a(\$a, \$c = '#\(([^()]*)\)#e', \$d = 'a("\$1", "#^ *-?[\d.]+ *\S *-?[\d.]+ *#e", "\\$0")') {
\$e = 'preg_replace';
while (\$a != \$b = \$e(\$c, \$d, \$a)) {
\$a = \$b;
}
return \$e('#^(.*)\$#e', \$d, \$a);
}
``````

Tests:

``````assert(a('1 + 3 / -8') === '-0.5');
assert(a('2*3*4*5+99') === '219');
assert(a('4 * (9 - 4) / (2 * 6 - 2) + 8') === '10');
assert(a('1 + ((123 * 3 - 69) / 100)') === '4');
assert(a('2.45/8.5*9.27+(5*0.0023)') === '2.68344117647');
assert(a(' 2 * 3 * 4 * 5 + 99 ') === '219');
``````
The e modifier to preg_replace violates the rule that evals are forbidden.
A:

Is there any code with optimization? Such as multiplication on zero or serial infix minus. On Python or C.

A:

OCaml using Camlp4 directly:

``````open Camlp4.PreCast

let expr = Gram.Entry.mk "expr"

EXTEND Gram
expr:
[   [ e1 = expr; "+"; e2 = expr -> e1 + e2
| e1 = expr; "-"; e2 = expr -> e1 - e2 ]
|   [ e1 = expr; "*"; e2 = expr -> e1 * e2
| e1 = expr; "/"; e2 = expr -> e1 / e2 ]
|   [ n = INT -> int_of_string n
| "("; e = expr; ")" -> e ]   ];
END

let () = Gram.parse expr Loc.ghost (Stream.of_string "1-2+3*4")
``````
``````open Genlex

let lex = make_lexer ["+"; "-"; "*"; "/"; "("; ")"]

let rec parse_atom = parser
| [< 'Int n >] -> n
| [< 'Kwd "("; e=parse_expr; 'Kwd ")" >] -> e
and parse_factor = parser
| [< e1=parse_atom; stream >] ->
(parser
| [< 'Kwd "*"; e2=parse_factor >] -> e1 * e2
| [< 'Kwd "/"; e2=parse_factor >] -> e1 / e2
| [< >] -> e1) stream
and parse_expr = parser
| [< e1=parse_factor; stream >] ->
(parser
| [< 'Kwd "+"; e2=parse_expr >] -> e1 + e2
| [< 'Kwd "-"; e2=parse_expr >] -> e1 - e2
| [< >] -> e1) stream

let () =
Printf.printf "%d\n" (parse_expr(lex(Stream.of_string "1 + 2 * (3 + 4)")));;
``````