C recursion - increase each digit of a number by 1 except digit number 9 which will be 0

Question

i already managed to create a function which returns every number's digit(0-8) by 1.

example : input : 3 output: 4

input : 345 output: 456

But i have a problem finding a solution for digit number 9 which need to return 0 .

example: input : 9 output: 0

input : 945 output: 56

input : 99 output: 0

input : 19 output: 20

NOTE: Do not specifically check if the digit is 9

my code:

int new_num(int num){
    int dig = num%10;
    num = num/10;

    if( num==0 ){
        return dig+1;
    }
    int res = new_num(num);
    dig +=1;
    res *=10;
    res +=dig;
    return res;   
}

thanks.

Answer 1

Here you are.

#include <stdio.h>

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

    return ( n % Base + 1 ) % Base +
           ( n / Base == 0 ? 0
                             : Base * new_num( n / Base ) );
}

int main(void) 
{
    printf( "%d -> %d\n", 99, new_num( 99 ) );
    printf( "%d -> %d\n", 945, new_num( 945 ) );
    printf( "%d -> %d\n", 19, new_num( 19 ) );
    printf( "%d -> %d\n", 123456789, new_num( 123456789 ) );

    return 0;
}

The program output is

99 -> 0
945 -> 56
19 -> 20
123456789 -> 234567890