CODEWITHSUNDEEP
cexceptionfloating-point
Vandana Tummala
Vandana Tummala

Reputation: 1

floating point exception in c program

while running this code I am getting floating point exception pls explain why it is coming so.

#include <stdio.h>
int gcd(long int, int);
int main() {
    int t, n, a, i;
    long int abc;
    scanf("%d", &t);
    while (t--) {
        abc = 1;
        scanf("%d", &n);
        abc = n * (n - 1);
        for (i = n - 2; i > 1; i--) {
            a = gcd(abc, i);
            abc = ((abc * i) / a);
        }
        printf("%ld \n", abc);
    }
    return 0;
}
int gcd(long int a, int b) {
    if (b == 0)
        return a;
    else {
        return (b, a % b);
    }
 }

Upvotes: 0

Views: 336

Answers (2)

meernet
meernet

Reputation: 386

You should change a few things:

1) if(a == 0) instead of b==0 and return variable b
2) Use recursion gcd(b%a, a)

if (a == 0)
       return b;
    else {
        return gcd(b%a, a);
    }

And you can also use this short version

return  a ? gcd(b%a,a) : b;

Proof:

We have m = n * (m div n) + m Mod n.

This equation is correct even if n = 1, because m div 1 = m and m mod 1 =0.

If n and m mod n it's multiple of d, also m is multiple of d, because n * (m div n) and m mod n / d.

On the other hand if m and n is multiple of e, to because m mod n = m -n * (m div n) is also multiple of e.

Upvotes: 0

Miroslav Franc
Miroslav Franc

Reputation: 1352

The else part in gcd function is bogus. You probably wanted to call gcd recursively and instead you are returning a % b. And as a result if a % b == 0 you divide by 0 on line 13.

The expression (b, a % b) is evaluated as two subexpressions separated by comma operator. The value of b is forgotten and the value of the whole expression becomes a % b.

The correct version:

int gcd(long int a, int b) {
    if (b == 0)
        return a; 
    else {
        return gcd(b, a % b);
    }
 }

Upvotes: 4

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