I just wrote a simple function to convert a number with any base into a base 10 number:

#include <stdio.h>
#include <math.h>
#include <string.h>

int tobase10(int *input, int size, int base);

int main() {
	int input[256] = {2, 2, 1, 0};

	printf("%d\n", tobase10(input, 4, 3));

	printf("Press any key to continue ...\n");
	getchar();
	return 0;}

int tobase10(int *input, int size, int base) {
	int result = 0;
	int count = 0;
	
	while((size+1) != 0) {
		result = (input[count] * pow(base, (size-1))) + result;
		count++;
		size--; }
	return result; }

I now want to modify it so it edits the input array to put every digit of the number in separate array block. Then I want it to return the number of digets there are. The only way I can think of doing this is puting the result in a string, use strlen() to get the size, and use a loop with atoi() to read it back into the input array. Is there any better, more elegant way to do this?
It seems to get ugly when I try.

(int)log10(n) + 1 will give you the number of digits.
You could strip digits off with a loop body like this:

digit = n % 10;
// store digit or whatever
n /= 10;

There are two generally accepted ways to split a number into digits. First is the string conversion:

#include <stdio.h>

int main ( void )
{
    int value;

    printf ( "Enter an integer: " );
    fflush ( stdout );

    if ( scanf ( "%d", &value ) == 1 ) {
        char buffer[50]; /* Arbitrary "large enough" size */
        size_t i;

        /* Quick and easy, but relatively expensive */
        sprintf ( buffer, "%d", value );

        for ( i = 0; buffer[i] != '\0'; i++ )
            printf ( "%c\n", buffer[i] );
    }

    return 0;
}

Next is the mathematical solution, where you break the number down directly:

#include <stdio.h>

int main ( void )
{
    int value;

    printf ( "Enter an integer: " );
    fflush ( stdout );

    if ( scanf ( "%d", &value ) == 1 ) {
        if ( value == 0 ) {
            /* A typical breakdown loop doesn't handle 0 */
            printf ( "0\n" );
        }
        else {
            /* This technique works for any base */
            const int radix = 10;

            /*
              Note that the digits come in least 
              to most significant order 
            */
            while ( value != 0 ) {
                printf ( "%d\n", value % radix );
                value /= radix;
            }
        }
    }

    return 0;
}

Most solutions are variations of these two general concepts.

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