Hi,
Simple question - In c++, what's the neatest way of getting which of two numbers (u0 and u1) is the smallest positive number? (that's still efficient)
Every way I try it involves big if statements or complicated conditional statements.
Thanks, Dan
Here's a simple example:
bool lowestPositive(int a, int b, int& result)
{
//checking code
result = b;
return true;
}
lowestPositive(5, 6, result);
I suggest you refactor the function into simpler functions. Furthermore, this allows your compiler to better enforce expected input data (though, what I've coded should produce a warning in the call to smallerInt)
unsigned int smallerInt( unsigned int a, unsigned int b )
{
return ( a < b ) ? a : b;
}
bool lowestPositive( int a, int b, int& result )
{
if ( a < 0 || b < 0 )
{
return false;
}
result = smallerInt( a, b );
return true;
}
If the values are represented in twos complement, then
result = ((unsigned )a < (unsigned )b) ? a : b;
will work since negative values in twos complement are larger, when treated as unsigned, than positive values. As with Jeff's answer, this assumes at least one of the values is positive.
return result >= 0;
Three lines with the use (abuse?) of the ternary operator
int *smallest_positive(int *u1, int *u2) {
if (*u1 < 0) return *u2 >= 0 ? u2 : NULL;
if (*u2 < 0) return u1;
return *u1 < *u2 ? u1 : u2;
}
Don't know about efficiency or what to do if both u1 and u2 are negative. I opted to return NULL (which has to be checked in the caller); a return of a pointer to a static -1 might be more useful.
Edited to reflect the changes in the original question :)
bool smallest_positive(int u1, int u2, int& result) {
if (u1 < 0) {
if (u2 < 0) return false; /* result unchanged */
result = u2;
} else {
if (u2 < 0) result = u1;
else result = u1 < u2 ? u1 : u2;
}
return true;
}
unsigned int mask = 1 << 31;
unsigned int m = mask;
while ((a & m) == (b & m)) {
m >>= 1;
}
result = (a & m) ? b : a;
return ! ((a & mask) && (b & mask));
EDIT: Thought this is not so interesting so I deleted it. But on the second thought, just leave it here for fun :) This can be considered as a dump version of Doug's answer :)
Hack using "magic constant" -1:
enum
{
INVALID_POSITIVE = -1
};
int lowestPositive(int a, int b)
{
return (a>=0 ? ( b>=0 ? (b > a ? a : b ) : INVALID_POSITIVE ) : INVALID_POSITIVE );
}
This makes no assumptions about the numbers being positive.
Pseudocode because I have no compiler on hand:
////0 if both negative, 1 if u0 positive, 2 if u1 positive, 3 if both positive
switch((u0 > 0 ? 1 : 0) + (u1 > 0 ? 2 : 0)) {
case 0:
return false; //Note that this leaves the result value undef.
case 1:
result = u0;
return true;
case 2:
result = u1;
return true;
case 3:
result = (u0 < u1 ? u0 : u1);
return true;
default: //undefined and probably impossible condition
return false;
}
This is compact without a lot of if statements, but relies on the ternary " ? : " operator, which is just a compact if, then, else statement. "(true ? "yes" : "no")" returns "yes", "(false ? "yes" : "no") returns "no".
In a normal switch statement after every case you should have a break;, to exit the switch. In this case we have a return statement, so we're exiting the entire function.
Might get me modded down, but just for kicks, here is the result without any comparisons, because comparisons are for whimps. :-)
bool lowestPositive(int u, int v, int& result)
{
result = (u + v - abs(u - v))/2;
return (bool) result - (u + v + abs(u - v)) / 2;
}
Note: Fails if (u + v) > max_int. At least one number must be positive for the return code to be correct. Also kudos to polythinker's solution :)
My idea is based on using min and max. And categorized the result into three cases, where
The best thing is that it's not look too complicated. Code:
bool lowestPositive(int a, int b, int& result)
{
int min = (a < b) ? a : b;
int max = (a > b) ? a : b;
bool smin = min > 0;
bool smax = max > 0;
if(!smax) return false;
if(smin) result = min;
else result = max;
return true;
}
I prefer clarity over compactness:
bool lowestPositive( int a, int b, int& result )
{
if (a > 0 && a <= b) // a is positive and smaller than or equal to b
result = a;
else if (b > 0) // b is positive and either smaller than a or a is negative
result = b;
else
result = -1; // neither are positive
return result > 0; // zero is not positive
}
This will handle all possible inputs as you request.
bool lowestPositive(int a, int b, int& result)
{
if ( a < 0 and b < 0 )
return false
result = std::min<unsigned int>( a, b );
return true;
}
That being said, the signature you supply allows sneaky bugs to appear, as it is easy to ignore the return value of this function or not even remember that there is a return value that has to be checked to know if the result is correct.
You may prefer one of these alternatives that makes it harder to overlook that a success result has to be checked:
boost::optional<int> lowestPositive(int a, int b)
{
boost::optional<int> result;
if ( a >= 0 or b >= 0 )
result = std::min<unsigned int>( a, b );
return result;
}
or
void lowestPositive(int a, int b, int& result, bool &success)
{
success = ( a >= 0 or b >= 0 )
if ( success )
result = std::min<unsigned int>( a, b );
}
tons of the answers here are ignoring the fact that zero isn't positive :)
with tricky casting and tern:
bool leastPositive(int a, int b, int& result) {
result = ((unsigned) a < (unsigned) b) ? a : b;
return result > 0;
}
less cute:
bool leastPositive(int a, int b, int& result) {
if(a > 0 && b > 0)
result = a < b ? a : b;
else
result = a > b ? a : b:
return result > 0;
}
With all due respect, your problem may be that the English phrase used to describe the problem really does hide some complexity (or at least some unresolved questions). In my experience, this is a common source of bugs and/or unfulfilled expectations in the "real world" as well. Here are some of the issues I observed:
Some programmers use a naming
convention in which a leading u
implies unsigned, but you didn't
state explicitly whether your
"numbers" are unsigned or signed
(or, for that matter, whether they
are even supposed to be integral!)
I suspect that all of us who read it assumed that if one argument is positive and the other is not, then the (only) positive argument value is the correct response, but that is not explicitly stated.
The description also doesn't define the required behavior if both values are non-positive.
Finally, some of the responses offered prior to this post seem to imply that the responder thought (mistakenly) that 0 is positive! A more specific requirements statement might help prevent any misunderstanding (or make it clear that the issue of zero hadn't been thought out completely when the requirement was written).
I'm not trying to be overly critical; I'm just suggesting that a more precisely-written requirement will probably help, and will probably also make it clear whether some of the complexity you're concerned about in the implementation is really implicit in the nature of the problem.
Here's a fast solution in C using bit twiddling to find min(x, y). It is a modified version of @Doug Currie's answer and inspired by the answer to the Find the Minimum Positive Value question:
bool lowestPositive(int a, int b, int* pout)
{
/* exclude zero, make a negative number to be larger any positive number */
unsigned x = (a - 1), y = (b - 1);
/* min(x, y) + 1 */
*pout = y + ((x - y) & -(x < y)) + 1;
return *pout > 0;
}
Example:
/** gcc -std=c99 *.c && a */
#include <assert.h>
#include <limits.h>
#include <stdio.h>
#include <stdbool.h>
void T(int a, int b)
{
int result = 0;
printf("%d %d ", a, b);
if (lowestPositive(a, b, &result))
printf(": %d\n", result);
else
printf(" are not positive\n");
}
int main(int argc, char *argv[])
{
T(5, 6);
T(6, 5);
T(6, -1);
T(-1, -2);
T(INT_MIN, INT_MAX);
T(INT_MIN, INT_MIN);
T(INT_MAX, INT_MIN);
T(0, -1);
T(0, INT_MIN);
T(-1, 0);
T(INT_MIN, 0);
T(INT_MAX, 0);
T(0, INT_MAX);
T(0, 0);
return 0;
}
Output:
5 6 : 5
6 5 : 5
6 -1 : 6
-1 -2 are not positive
-2147483648 2147483647 : 2147483647
-2147483648 -2147483648 are not positive
2147483647 -2147483648 : 2147483647
0 -1 are not positive
0 -2147483648 are not positive
-1 0 are not positive
-2147483648 0 are not positive
2147483647 0 : 2147483647
0 2147483647 : 2147483647
0 0 are not positive
After my first post was rejected, allow me to suggest that you are prematurely optimizing the problem and you shouldn't worry about having lots of if statements. The code you're writing naturally requires multiple 'if' statements, and whether they are expressed with the ternary if operator (A ? B : C) or classic if blocks, the execution time is the same, the compiler is going to optimize almost all of the code posted into very nearly the same logic.
Concern yourself with the readability and reliability of your code rather than trying to outwit your future self or anyone else who reads the code. Every solution posted is O(1) from what I can tell, that is, every single solution will contribute insignificantly to the performance of your code.
I would like to suggest that this post be tagged "premature optimization," the poster is not looking for elegant code.
uint lowestPos(uint a, uint b) { return (a < b ? a : b); }
You are looking for the smallest positive, it is be wise to accept positive values only in that case. You don't have to catch the negative values problem in your function, you should solve it at an earlier point in the caller function. For the same reason I left the boolean oit.
A precondition is that they are not equal, you would use it like this in that way:
if (a == b)
cout << "equal";
else
{
uint lowest = lowestPos(a, b);
cout << (lowest == a ? "a is lowest" : "b is lowest");
}
You can introduce const when you want to prevent changes or references if you want to change the result. Under normal conditions the computer will optimize and even inline the function.