Bit Reversal in an 8-bit Value

Question

Problem: Bit Reversal in an 8-bit Value. Solve this hands-on electronics task on EWskills.

Accepted Answer

#include 
#include 

// Reverse bits in an 8-bit value
uint8_t reverse_bits(uint8_t val) {
    uint8_t result = 0;

    for (int i = 0; i < 8; i++) {
        // Shift result left to make room for next bit
        result <<= 1;

        // Take LSB from input and add to result
        result |= (val & 1);

        // Shift input right to process next bit
        val >>= 1;
    }

    return result;
}

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

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

Why Bit Reversal?

Used in:

CRC and parity computation
Bitstream processing
SPI/UART LSB-first data handling

Logic Summary

Initialize result = 0
Loop 8 times:
- Shift result left by 1
- Copy the last bit of val to result
- Shift val right
End result is val reversed bit by bit

36. Bit Reversal in an 8-bit Value