Bit Reversal in an 8-bit Value

Code

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

uint8_t reverse_bits(uint8_t val) {
    // reverse nibbles
    val = ((val & 0xF0) >> 4) | ((val & 0x0F) << 4); 

    // reverse 2 bits
    val = ((val & 0xCC) >> 2) | ((val & 0x33) << 2);

    // reverse individual bits
    val = ((val & 0xAA) >> 1) | ((val & 0x55) << 1);

    return val;
}

int main() {
    uint8_t val;
    scanf("%hhu", &val);

    uint8_t result = reverse_bits(val);
    printf("%u", result);
    return 0;
}

/*

i = 0001 1010
o = 0101 1000

swap nibbles (4 bytes) (mask = 0xF0 for left nibble and 0x0F for right nibble)
1010 0001

swap 2 bits (mask = 0b1100 1100 (0xCC) for left sides and 0b0011 0011 (0x33) for right sides)
1010 0100

swap 1 bits  (mask = 0b1010 1010 (0xAA) for left sides and 0b0101 0101 (0x55) for right sides)
0101 1000    

*/

Solving Approach