How do I create a function that will take two integers, num1, num2, and return an integer and the function will check if the numeric value of the number num2 is in the digits of the number num1. example num1 = 51749, num2 = 566: num1 have 17, num2 sum is 17

int func(int num1, int num2) this is the start of the function with no array. I started this one but for some reason it's not working can you please explain me what is the problem.

#include <stdio.h>
#include <stdlib.h>

int func(int num1, int num2) {
    int sum = 0, sum2 = 0, ad = 0;
    int digit;

    while (num1 > 0) {
         digit = num1 % 10;
         sum = sum   digit;
         num1 = num1 / 10;
    }
    while (num2 > 0) {
         digit = num2 % 10;
         sum2 = sum2   digit;
         num2 = num2 / 10;
    }
    while (ad < 10) {
        if (sum =! num2)
            printf("The number is correct");
        ad  ;
    }
    printf("Different");
}

int main() {
    int a, b;
    printf("Type the first number ");
    scanf("%d", &a);
    printf("Type the second number ");
    scanf("%d", &b);
    if (func(a, b) == 1)
        printf("Yes");
    if (func(a, b) == 0)
        printf("No");

    return 0;
}

CodePudding user response：

There is a typo in if (sum =! num2): you probably mean this instead:

if (sum != num2)

In case you wonder why the bogus expression compiles at all, sum =! num2 is parsed as sum = !num2 which assigns 0 or 1 to sum if num2 is zero or not.

Note also that your function does not test whether the sum of digits in num2 exists inside num1. You must use a different method:

• compute sum2: the sum of digits in num2
• test if this number is present in num1 either as a 1-digit or a 2-digit number (because sum2 has either 1 or 2 digits).
• you may need to make a special case of (0, 0).

Here is a modified version:

#include <stdio.h>

int func(int num1, int num2) {
    int sum2 = 0;

    while (num2 > 0) {
         sum2  = num2 % 10;
         num2 /= 10;
    }
    if (sum2 == 0 && num1 == 0)
        return 1;   // special case for (0, 0)

    while (num1 > 0) {
         if (num1 % 10 == sum2)
             return 1;   // sum2 has 1 digit and is present
         if (num1 % 100 == sum2)
             return 1;   // sum2 has 2 digits and is present
         num1 /= 10;
    }
    return 0;  // sum2 is not present
}

int main() {
    int a = 0, b = 0;
    printf("Type the first number: ");
    scanf("%d", &a);
    printf("Type the second number: ");
    scanf("%d", &b);
    if (func(a, b)) {
        printf("Yes\nsum of digits in %d is present in %d\n", b, a);
    } else {
        printf("No\nsum of digits in %d is not present in %d\n", b, a);
    }
    return 0;
}