c - how to get the intended result when the second input b equals 0?

Question

When I ran this code and provided 28000/3 as input, it showed:

28000/32766 = 0.

Why on earth would this happen? I'm new to c and this is really confusing.

#include <stdio.h>

int main(void) {
    int divide(int a, int b, int *result);
    int a, b;
    
    scanf("%d %d", &a, &b);
    
    int c;
    
    if (divide(a, b, &c)) {
        printf("%d/%d=%d
", a, b, c);
    }
    
    return 0;
}

int divide(int a, int b, int *result) {
    int ret = 1;
    if (b == 0) 
        ret = 0;
    else {
        *result = a / b;
    }
    return ret;
}

question from:https://stackoverflow.com/questions/66058117/how-to-get-the-intended-result-when-the-second-input-b-equals-0

Answer 1

The scanf() doesn't know if you would enter a division sign /. You need to change its format:

scanf("%d/%d", &a, &b);

So that you could enter 28000/3 or similar inputs.

---

Here's the perfect code (notice comments):

#include <stdio.h>
#include <limits.h>

// Definitions must be global
int divide(int a, int b, int *result);

int main(void) {
    int a, b;

    // It's programmer's responsibility to ensure the input
    if (scanf("%d/%d", &a, &b) != 2) {
        puts("Arguments are incorrectly passed.");
        return -1;
    }

    int c;

    if (divide(a, b, &c))
        printf("%d/%d=%d
", a, b, c);
    else
        printf("b is zero.
"); // also code the 'else' to print an error
                                // if b is zero

    return 0;
}

int divide(int a, int b, int *result) {
    int ret = 1;
    if (b == 0 || (a == INT_MIN && b == -1))
        ret = 0;
    else
        *result = a / b;
    
    return ret;
}