如何将整数分成多个数字？

Question

I have a remainder function that finds that modulo of a number then a divider function to divide that number. My program isn't working the way I need it to. For example if I put in 2502 as my number, I should get the output of : 2 5 0 2. I need to be able to store the value through each iteration, so for example:

number: 123
    123 % 10 = 3  //last digit
Number: 123 / 10 = 12
    12  % 10 = 2  //second digit
Number: 12  / 10 = 1
    1   % 10 = 1  //first digit

---

int Rem(int num);
int Div(int num);

int main() {
    int num;
    printf("Enter an integer between 1 and 32767: ");
    scanf("%d", &num);
    Rem(num);
    Div(num);
    printf("%d","The digits in the number are: ");

}


    int Rem(int num) {
        while(num != 0){
        int rem = num % 10;
        return rem;
        }
    }

    int Div(int num){
        while(num != 0){
        int div = num / 10;
        return div;
        }
    }

Answer 1

The idea here is pretty simple, but there are some subtleties. Here's some pseudo code. You'll need to convert to C.

num = 9934; // or get it from input
do {
    rem = num % 10;  // this gives you the lowest digit
    num = num / 10;  // divide by 10 to get rid of that lowest digit
    print rem; 
} while (num != 0);

I use a do ... while loop so the output will be correct if the user enters 0.

If you code this up and run it, you'll notice that it prints the digits in reverse order: 4 3 9 9 . So you'll need some way to reverse the digits before you output them. Three possible ways are:

1. Store the digits in an array, and then reverse the array before outputting.
2. Push each digit onto a stack. When you're done, pop each digit off the stack and output it.
3. Write a recursive function. That would eliminate the need for an explicit stack or array.

Answer 2

您也可以将其转换为字符串，然后再转换回该字符串的元素，但请记住“ \\ 0”

Answer 3

You can simply use this function , it return the number of digits passing by argument

size_t  ft_cnbr(int n)
    {
        size_t count;

        count = 1;
        if (n < 0)
        {
            count++;
            n = -n;
        }
        while (n > 9)
        {
            n = n / 10;
            count++;
        }
        return (count);
    }