Bit Reversal in an 8-bit Value

Code

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

uint8_t reverse_bits(uint8_t val) {
    // Your logic here
    // 0b1010 -> 0b0101
    uint8_t rev = 0;
    for(int i = 0; i < 8; i++){
        rev |= (((val >> (7 - i)) & 1 ) << i);
    }

    return rev;
}

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

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

Solving Approach