do this using recursion only

Question

Write a recursive function evenDigits that accepts an integer parameter n and returns a new integer containing only the even digits from n, in the same order. If n does not contain any even digits, return 0.

For example, the call of evenDigits(8342116) should return 8426 and the call of evenDigits(35179) should return 0.

Answer 1

Here are my five cents.:)

#include <stdio.h>

int evenDigits( int n )
{
    const int Base = 10;

    int digit = n % Base;

    int even = digit % 2 == 0;

    if ( !even ) digit = 0;

    return ( n /= Base ) == 0 ? digit 
                              : ( even ? Base : 1 ) * evenDigits( n ) + digit;
}

int main(void) 
{
    printf( "evenDigits( 8342116 ) = %d\n", evenDigits( 8342116 ) );
    printf( "evenDigits( -8342116 ) = %d\n", evenDigits( -8342116 ) );
    printf( "evenDigits( 35179 ) = %d\n", evenDigits( 35179 ) );
    printf( "evenDigits( -35179 ) = %d\n", evenDigits( -35179 ) );

    return 0;
}

The program output is

evenDigits( 8342116 ) = 8426
evenDigits( -8342116 ) = -8426
evenDigits( 35179 ) = 0
evenDigits( -35179 ) = 0