Bit Reversal in an 8-bit Value

Code

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

uint8_t reverse_bits(uint8_t val) {
    // Your logic here
    /*
    val = (val << 1) | (val >> 7);
    val = ((((((val << 1) & (~((1 << 2) - 1))) >> 1) | (val >> 7)) << 1) | (val & 1));
    val = ((((((val << 1) & (~((1 << 3) - 1))) >> 2) | (val >> 7)) << 2) | (val & 3));
    val = ((((((val << 1) & (~((1 << 4) - 1))) >> 3) | (val >> 7)) << 3) | (val & 7));
    val = ((((((val << 1) & (~((1 << 5) - 1))) >> 4) | (val >> 7)) << 4) | (val & 15));
    val = ((((((val << 1) & (~((1 << 6) - 1))) >> 5) | (val >> 7)) << 5) | (val & 31));
    val = ((((((val << 1) & (~((1 << 7) - 1))) >> 6) | (val >> 7)) << 6) | (val & 63));
    */

    return __builtin_bitreverse8(val);
}

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

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

Solving Approach