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 reg=0;
    uint8_t bit;

    for(int i=0; i<=7; i++){
        bit=(val&0x1);
        //printf("%u\n",bit);
        reg|=(bit<<7-i);

       // printf("%u ",reg);
        val=val>>1;

    }


    return reg;
}

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

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

Solving Approach