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 = 0x00;
    for(uint8_t i=0; i<8; i++){
        if((val>>i)&1){
            reg |= (1<<(7-i));
        }
    }
    return reg;
}

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

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

Solving Approach

 

 

 

Upvote
Downvote
Loading...

Input

26

Expected Output

88