Bit Reversal in an 8-bit Value

Code

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

uint8_t reverse_bits(uint8_t val) {
    // Your logic here
    uint8_t rev_val = 0;
    uint8_t buffer = 0;

    for (int i=0; i<8; i++) {
        buffer = (val >> i) & 1;
        rev_val |= buffer << (7-i);
    }
    return rev_val;
}

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

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

Solving Approach