Converting from decimal to base 4 using bitmasking

Question

I'm trying to write a program that converts from decimal to base 4 using bit masking. First I wrote this program that converts from decimal to binary to understand how it works

#include <stdio.h>
void binary(unsigned int num) {   
    int i;                                                                          
    unsigned temp, mask=1;                    
    mask<<=31;                       
    for(i=0; i<32; i++) {     
        temp=num&mask;                  
        printf("%d", temp>>(31-i));
        mask>>=1;
    }
}
int main () {
    int n;
    printf("Enter a number:\n");
    scanf("%d", &n);
    binary(n);
    return 0;
}

Then I was trying to use the same approach here

void base4(unsigned int num) {   
    int i;                                                                          
    unsigned temp, mask=3;                    
    mask<<=15;                       
    for(i=0; i<16; i++) {     
        temp=   // <<<==== ???               
        printf("%d", temp>>(15-i)*2);
        mask>>=1;
    }

I know that in in base 4, numbers use 2 bits. So if we are ANDing each bit of a given value lets say 22 with the corresponding bit of the mask=3 we will end up with something like this

.... 0  0  1  1  2  (base4)
.... 00 00 01 01 10 (bin)

I just couldn't use this information to apply it to my previous approach, would be great if someone could help.

Answer 1

char *toBase4(char *buff, unsigned val, int printZeroes)
{
    char *wrk = buff;
    int shift = sizeof(val) * CHAR_BIT - 2;
    unsigned mask = 3 << shift;

    do
    {
        if((val & mask) || printZeroes) 
        {
            *wrk++ = ((val & mask) >> shift) + '0';
            printZeroes = 1;
        }
        mask >>= 2;
    }while((shift -= 2) >= 0);
    *wrk = 0;
    return buff;
}

int main(void)
{
    char x[256];

    printf("%s\n", toBase4(x, 0xff00, 0));
    printf("%s\n", toBase4(x, 0xff00, 1));
}