Print all the possible strings a telephone number can represent

The below code doesn't take into account the fact that keys 7 and 9 represent 4 alphabets and also it just prints ' ' and '+' for key 1 and 0 respectively. The code is just to explain the logic for printing the combinations and additional logic can be added to handle these cases.

Code:
 #include<stdio.h>
#include<stdlib.h>
#include<string.h>

char digitToChar(char x, int index)
{
    char ch;

    switch(x)
    {
        case '0':
            ch = '+'; break;
        case '1':
            ch = ' '; break;
        case '2':
            ch = "abc"[index]; break;
        case '3':
            ch = "def"[index]; break;
        case '4':
            ch = "ghi"[index]; break;
        case '5':
            ch = "jkl"[index]; break;
        case '6':
            ch = "mno"[index]; break;
        case '7':
            ch = "pqrs"[index]; break;
        case '8':
            ch = "tuv"[index]; break;
        case '9':
            ch = "wxyz"[index]; break;
        default:
            ch = ' ';

    }
    return ch;
}

void printAll(char* num, char* arr, int len, int x)
{
    int i=0;

    if(x == len)
    {
        printf("%s\n", arr);
        return;
    }

    for(i=0; i<3; i++)
    {
        arr[x] = digitToChar(num[x], i);
        printAll(num, arr, len, x+1);
    }
}

int main()
{
    char* num = "8017393450";
    char arr[20] = {0,};

    printAll(num, arr, 10, 0);

    return 0;
}