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\n", 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;
}

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\n", a, b, c);
    else
        printf("b is zero.\n"); // 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;
}

Answer 2

Apart from the wrong input for your scanf (see other answer and comments),
you also have to different return value concepts in your function divide().
You are using both the return value AND the return by pointer parameter, but inconsistently in the two branches of the check against 0.

To make it consistent change the function like:

#include<stdio.h>
int main(void)
{
    int divide(int a,int b,int *result);
     int a=2021,b=42;
     if(2!=scanf("%d %d",&a,&b))
     {
         printf("Scanf failed!\n");
         return 0;
     }
     int c=1970;
     if(!divide(a,b,&c))
     {
         printf("Division failed!\n");
     } else 
     {
         printf("%d/%d=%d\n",a,b,c);
     }
     
     return 0;
}
int divide(int a,int b,int *result)
{
    int ret=1;
    if(b==0)
    {
        ret=0;
        *result = 0;
    } else
    {
        *result=a/b;
        ret = 1; /* not 100% necessary because of init; for symmetry */
    }
    return ret;
}

That gets you (with input "32 4"):

32/4=8

With input "32 0":

Division failed!

With input "32/0":

Scanf failed!

Answer 3

There are multiple problems in your code:

• [Major] you do not test that scanf() successfully converted 2 integers into a and b. When you type 28000/3, the conversion specification "%d %d" causes scanf() to properly convert 28000 into a but the / does not start a number so scanf() stops there and returns 1, leaving b unchanged. b is uninitialized and happens to contain the value 32766 which explains the output 28000/32766 = 0. You should change the format and test the return value.
• [Major] you do not update *result in divide() in case of overflow, so c would be unchanged and still uninitialized if b was 0.
• [Minor] forward declarations such as int divide(int a, int b, int *result); should appear at the global scope. The compiler accepts them in a local scope but it is considered bad style and confusing. The reason for this is the declaration only extends to the end of the main() function, so the actual definition of the divide() function could have a different prototype and this inconsistency would go undetected.
• [Minor] integer division can overflow and cause undefined behavior (ie: program termination) if b=0, but also if a=INT_MIN (-2147483648) and b=-1. You should detect and report this overflow to the caller using the return value so the caller can distinguish division overflow from a regular 0 quotient.

Here is a modified version:

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

/* forward declaration at global scope */
int divide(int a, int b, int *result);

int main(void) {
    int a, b, c;
 
    /* read input values and test for success */
    if (scanf("%d/%d", &a, &b) != 2) {
        printf("invalid input\n");
        return 1;
    }
    
    /* perform division and test for overflow */
    if (divide(a, b, &c)) {
        printf("division overflow: %d/%d\n", a, b);
        return 1;
    } else {
        printf("%d/%d=%d\n", a, b, c);
        return 0;
    }
}

int divide(int a, int b, int *result) {
    /* test all cases of division overflow */
    if (b == 0 || (a == INT_MIN && b == -1)) {
        *result = 0;
        return -1;
    } else {
        *result = a / b;
        return 0;
    }
}