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 reversal = 0 ;
    uint8_t bit = 0;
for(uint8_t i = 0; i < 8; i++)
{
    bit = (val>>i) & 0x01;
    reversal = reversal << 1;
    reversal |= bit;
}

    return reversal;
}

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

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

Solving Approach