why n mod 10 in loop doesn't show output

Question

I have solve a basic problem in c that is count the digits in integer and I have written -

#include<stdio.h>

int main()   
{
    int n;

    scanf("%d",&n);
    
    int i;
    while(n!=0)
    {
        n %= 10;
        ++i;
    }
    
    printf("%d",i);
}

I already know that above code is wrong, I should write n/=10; instead of n%=10; but I wants to know why it is not printing even value of i i.e 0.

If I have written any wrong so please ignore it ,I am new here..

Because if `n` is not a multiple of `10`, then it will iterate forever. — Eugene Sh., Oct 26 '21 at 13:15
Why don't you print `n` inside the loop and observe what's happening? Also, `i` (uninitialized) is incremented forever - signed integer overflow causes UB. — babon, Oct 26 '21 at 13:19
Debug fail, as described by @babon. If you cannot debug, you cannot program computers. If nothing else, just put in some printf's so you can inspect important vars during thexrun, (or, of course, use an actual debugger:). — Martin James, Oct 26 '21 at 16:35

score 2 · Accepted Answer · answered Oct 26 '21 at 13:19

If the number n is not divisible by 10 then the value of this expression (the remainder of the division)

n %= 10;

will never be equal to 0.

Also the variable i is not initialized.

int i;

You should write

int i = 0;

do
{
    ++i;
} while ( n /= 10 );

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

1 Answers1