Question

I’d like the following code to work:

double num = 31415; /* num will never have a decimal part */
/* ... code here so that we can now say */
printf("%d", num_array[0] == 3 && num_array[1] == 1 && num_array[4] == 5); //1

I realize it’s trivial to do this with ints (just int i=0; while(num>0) numarr[size-++i] = num%10, num/=10; where size is predetermined to be the number of digits in the number), but that obviously won’t work for floats/doubles because you can’t mod one floats.

And yes, I need to use floats/doubles, despite not using the floating point section (it’s an exercise).

And I have figured out how to determine the number of digits in a double, using floor().

/* n > 0, n=floor(n) */
int numdigits(double n){
    int i = 0;
    while (n >0)
        n = floor(n/10), i++;
    return i;
}

Answer 1

Look up fmod. The number of digits will probably be easier to compute using log10.