c program for user to input two type of number and give output how many number user input

Question

User need to input 2 type of input 3 times, which is number 1 for pass and number 2 for fail...program need to give the summary of how many user input number 1 and number 2

#include <stdio.h>
int main() {
    int result[3];
    int pass;
    int fail;

    for (int i = 1; i <= 3; i++) {
        printf("Enter Result (Pass=1 Fail=2) :");
        scanf("%d", &result[i]);
    }
    if (&result == 1) {
        pass = pass + 1;
    }
    if (&result == 2) {
        fail = fail + 1;
    }
    printf("%d", pass);
    printf("%d", fail);
}

Answer 1

Some issues:

• &result == 1 takes the address of the array result and tries to compare that to the integer 1. This will, if the code even compiles, very rarely be true. You want to compare result[index] with 1 (or 2).
• You read all the inputs in a loop, but you only make the comparison once after the loop.
• Arrays are zero-based. Since result is an int[3], you are only allowed to access result[0], result[1] and result[2].
• Both pass and fail are uninitialized and contains indeterminate values. Reading those, which is required in pass = pass + 1;, will make the program have undefined behavior. Initialize them to 0.

I suggest not using an array and then do the comparisons with 1 and 2 inside the loop.

Example:

#include <stdio.h>

int main(void) {
    int result;
    int pass = 0;
    int fail = 0;

    for (int i = 0; i < 3; i++) {
        printf("Enter Result (Pass=1 Fail=2) :");

        if (scanf("%d", &result) != 1) {
            fputs("Erroneous input\n", stderr);
            return 1;
        }

        if (result == 1) {
            pass = pass + 1;
        } else if (result == 2) {
            fail = fail + 1;
        } else {
            printf("The invalid input %d will not be counted\n", result);
        }
    }

    printf("pass: %d\n", pass);
    printf("fail: %d\n", fail);
}