Count Set Bits in an 8-bit Register

Code

#include <stdio.h>
#include <stdint.h>

uint8_t count_set_bits(uint8_t reg) {
    // Your code here
    uint8_t count = 0;
    for(int i=0; i<8; i++){
        count += (reg>>i & 0x01);
    }
    return count;
}

int main() {
    uint8_t reg;
    scanf("%hhu", &reg);
    printf("%u", count_set_bits(reg));
    return 0;
}

Solving Approach

shift once  check the lsb is one , increment count and finally repeat 8 times.