C Golf: # spaces in string

Question

I'm trying to come up with with shortest C function (including spaces, any macros that are substituted into it) possible that is passed a NULL-terminated string and returns the number of spaces in that string. This is what I've got so far:

int n(char*l){return*l?!(*l-32)+n(++l):0;}

Which is 42 characters long. Any improvements?

Answer 1

38 characters:

n(char*l){return*l?!(*l-32)+n(++l):0;}

In c, the int return type declaration is superfluous.

Answer 2

#define meh return*l?!(*l-32)+n(++l):0

int n(char *l){meh;}

18 characters :P

Answer 3

40 - unfortunately, using (char*)1 as sentinel

m(l){return l-1?1+n(strchr(l,32)+1):-1;}

Answer 4

Obviously it's just for fun, so why limit it to C? :-)

APL

∇n x
1⊥x=' '
∇

For everybody who thinks APL is hard to read, note that you can convert this almost directly to longhand (i.e., Python):

def n(x):
    return sum(i==' ' for i in x)

Just say "del" instead of "def" and you're halfway there!

Answer 5

Haskell:

n=length.filter(==' ')

Answer 6

Clojure - 31 chars.

(def c #(count(re-seq #"\s"%)))

Exploded:

(defn count-spaces
  [astr] (count
           (re-seq #"\s" astr)))

count returns the number of items in the sequence, which, in this case, is the number of matches to the regex.

Answer 7

Golfscript - 10 chars

{" "/,(}:f

Here is the equivalent in python (Yes this is obviously not as short as using .count(' '))

f=lambda s:len(s.split(" "))-1

Iterating through the input counting spaces is one byte longer

{{32=},,}:f