The problem says that the user will input numbers till he inserts 0 (exit) then if there are any prime digits in that given number, the program shall multiply them.

For example: input : 4657 output : 35 ( 5 * 7 )

Here is what I have tried so far but I cannot pull it off with the multiplication... my code might look a little bit clumsy, I am a beginner :)

int main() {
    int number;
    int digit, prime;
    int i, aux;
    int multiplier;

  input:
    printf("Give Number :");
    scanf("%d", &number);
   
    do {
        multiplier = 1;
        digit = number % 10;
        number = number / 10;

        printf("%d \n", digit);

        prime = 1;
        for (i = 2; i <= sqrt(digit); i  ) {
            if (digit % 1 == 0) {
                prime = 0;
            }
        }
        if (prime != 0) {
            multiplier = multiplier * digit;
        }
    } while (number != 0);

    printf("The result  : %d\n", multiplier);

    goto input;

    return 0;
}

CodePudding user response：

There are multiple problems in your code:

• missing #include <stdio.h>
• use of label and goto statement to be avoided at this stage.
• testing for prime digits can be done explicitly instead of a broken loop.
• digit % 1 == 0 is always true, you should write digit % i == 0 (maybe a typo from copying someone else's code?
• i <= sqrt(digit) will not work unless you include <math.h> and still a bad idea. Use i * i < 10 instead.
• 0 and 1 should not be considered primes.
• it is unclear what the output should be if none of the digits are prime, let's assume 1 is OK.

Here is a modified version:

#include <stdio.h>

int main() {
    for (;;) {
        int number, multiplier;

        printf("Give Number :");
        if (scanf("%d", &number) != 1 || number <= 0)
            break;
   
        multiplier = 1;
        while (number != 0) {
            int digit = number % 10;
            number = number / 10;
            if (digit == 2 || digit == 3 || digit == 5 || digit == 7)
                multiplier *= digit;
        }
        printf("The result: %d\n", multiplier);
    }
    return 0;
}