Need math function (or algorithm) for reducing fractions

//Heres one Euclid came up with over 2 millennia ago
int gcd(int a,int b){
  while( b ){
    int c;
    c = a%b;
    a = b;
    b = c;
  }
  return a;
}

main(int argc,char *argv[]){
   int d,n,f;
   d = atoi(argv[1]);
   n = atoi(argv[2]);
   f = gcd(d,n);
   printf("%d/%d\n",d/f,n/f);
}

(If n was supposed to represent the numerator, I think I got my variable names backwards)

But it looks like Reduce(4,4) will only reduce to 2/2
and Reduce(4,-4) seems to have an even worse problem.

Why (q2guo) do you start at n/2? Shouldn't you check to see which is heigher?  What if the fraction is 10/2.? It will return 10/2 which can be 5/1...

Ready4Dis, I assumed that the fraction is a pure one, meaning that the numerator is always smaller that the denominator.

Then if n is assumed to be smaller than d, shouldn't you have started with d/2?
(I think that's what confused my labels:-) fortunately, the gcd approach doesn't depend such assumptions.

ozo: why would you start with d/2?  Since n is smaller than d
all i(s) between d/2 and n/2 are just extra work.

Given a fraction like 2/10, with the numerator 2 smaller than the denominator 10,
starting i=n/2 i.e. i=1 will fail to reduce it to 1/5
(with a fraction like 10/2, since n=10 and d=2, i=n/2 would start at 5, so Reduce(2,10) would find (10/2)/(2/2) = 5/1)

Ok, here is the function again

void Reduce(int &d, int &n) {
        int temp = (d >= n ? d : n);
        for (int i = temp; i > 1; i--) {
            if ( (n%i) == 0) {
                if ( (d%i) == 0) {
                     n = n / i;
                     d = d / i;
                     break;
                }
            }
        }
}

Still has a problem with -10/2
Here's my (Euclid's) function again with the variables renamed:

int gcd(int a,int b){
  while( b ){
    int c = a%b;
    a = b;
    b = c;
  }
  return a;
}

main(int argc,char *argv[]){
   int n,d,i;
   n = atoi(argv[1]);
   d = atoi(argv[2]);
   i = gcd(n,d);
   printf("%d/%d\n",n/i,d/i);
}