Thread: Would you check my code, please?

Hello, I'm new to C programming and would appreciate if someone would help with this.
Every time I run the following the program "stops" at the line where the gCD function takes in values (I've tried it without this line and it doesn't stop).

Code:

int abs(int a)
{
    int A=a;
    if (a<0) A=-a;
    return(A);
}



int gCD(int a, int b)
{
    int gcd = 1;

    while(a % 2 == 0 && b % 2 == 0)
    {
        gcd *= 2;
        a /= 2;
        b /= 2;
    }

    while (a % 2 == 0 || b % 2 == 0)
    {
        a /= 2;
        b /= 2;
    }

    int c = abs(a - b);
    while (c % 2 == 0)
    {
        c /= 2;
    }

    gcd *= c;

    return(gcd);
}





int main()
{
    int N;
    int D;

    printf("Type a numerator:  ");
    scanf("%d", &N);
    getchar();

    printf("\nType a denominator:  ");
    scanf("%d", &D);
    getchar();

    int gcdCN = gCD(N,D);

    N /= gcdCN;
    D /= gcdCN;

    printf("Simplified numerator:  %d\n", N);
    printf("Simplified denominator: %d\n\n", D);
    getchar();

    return 0;
}

Thanks!

It gets caught in an infinite loop depending upon your input, and even when it doesn't get caught, it returns the wrong answer. For example, with inputs N = 46, D = 146, it finishes, but it thinks the gcdCN is 50. It should be 2. An input that gets caught in the infinite loop is N = 2, D = 2. You might want to try again on your gcd algorithm and learn to use code tags.

Here's an algorithm in case you can not find one.
Euclidean algorithm - Wikipedia, the free encyclopedia

Note: I suggest changing

Code:

int gcdCN = gCD(N,D);

To two lines and adding the use of abs function to the call to prevent problems later for negative input.
Note: My code for Euclidean algorithm assumed non-negative input; your solution might not need abs on call.
Some C compilers will not like the declare and statement like you did it.

Code:

int gcdCN;

Code:

gcdCN = gCD(abs(N),abs(D));

Tim S.