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

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

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

Solving Approach