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 newVal = 0;
    for (uint8_t i = 0; i < 8; i++)
    {
        newVal |= ((val >> i) & 0x01) << (7 - i);
    }
    return newVal;
}

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

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

Solving Approach