Bit Reversal in an 8-bit Value

Code

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

uint8_t reverse_bits(uint8_t n) {
    // Your logic here
     n = ((n >> 1) & 0x55) | ((n & 0x55) << 1);

    // Step 2: swap consecutive pairs
    n = ((n >> 2) & 0x33) | ((n & 0x33) << 2);

    // Step 3: swap nibbles
    n = (n >> 4) | (n << 4);
    return n;
}

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

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

Solving Approach