Reverse Bits of a Number in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of reversing bits for 8-bit number

What is the output of reversing the bits of the 8-bit number 13 (00001101 in binary)?

DSA C

unsigned char reverseBits(unsigned char n) {
    unsigned char result = 0;
    for (int i = 0; i < 8; i++) {
        result <<= 1;
        result |= (n & 1);
        n >>= 1;
    }
    return result;
}

#include <stdio.h>

int main() {
    unsigned char num = 13;
    unsigned char reversed = reverseBits(num);
    printf("%u\n", reversed);
    return 0;
}

A 104

B 176

C 134

D 212

💡 Hint

Think about how bits move from right to left when reversed.

The binary of 13 is 00001101. Reversing bits gives 10110000 which is 176 in decimal.

❓ Predict Output

intermediate

Output of reversing bits for 16-bit number

What is the output of reversing the bits of the 16-bit number 4660 (0001001000110100 in binary)?

DSA C

unsigned short reverseBits16(unsigned short n) {
    unsigned short result = 0;
    for (int i = 0; i < 16; i++) {
        result <<= 1;
        result |= (n & 1);
        n >>= 1;
    }
    return result;
}

#include <stdio.h>

int main() {
    unsigned short num = 4660;
    unsigned short reversed = reverseBits16(num);
    printf("%u\n", reversed);
    return 0;
}

A 11336

B 13320

C 13312

D 13330

💡 Hint

Reverse the bits carefully from right to left for all 16 bits.

4660 in binary is 0001001000110100. Reversing bits gives 0010110001001000 which is 11336 in decimal.

🔧 Debug

advanced

Identify the error in bit reversal code

What error does the following code produce when reversing bits of an 8-bit number?

DSA C

unsigned char reverseBits(unsigned char n) {
    unsigned char result = 0;
    for (int i = 0; i < 8; i++) {
        result <<= 1;
        result |= (n & 1);
        n >>= 1;
    }
    return result;
}

A Compilation error due to loop condition

B Runs correctly and returns reversed bits

C Runs but returns incorrect result due to extra iteration

D Runtime error due to shifting beyond bit size

💡 Hint

Check the loop boundary condition carefully.

The loop runs 9 times (i <= 8), but should run 8 times (i < 8). The extra iteration shifts bits incorrectly causing wrong output.

❓ Predict Output

advanced

Output of reversing bits for 32-bit number

What is the output of reversing the bits of the 32-bit number 43261596 (00000010100101000001111010011100 in binary)?

DSA C

unsigned int reverseBits32(unsigned int n) {
    unsigned int result = 0;
    for (int i = 0; i < 32; i++) {
        result <<= 1;
        result |= (n & 1);
        n >>= 1;
    }
    return result;
}

#include <stdio.h>

int main() {
    unsigned int num = 43261596;
    unsigned int reversed = reverseBits32(num);
    printf("%u\n", reversed);
    return 0;
}

A 964176192

B 964176193

C 964176194

D 964176195

💡 Hint

Reverse all 32 bits carefully.

43261596 binary reversed is 00111001011110000010100101000000 which is 964176192 decimal.

🧠 Conceptual

expert

Why is bit reversal useful in algorithms?

Which of the following is the main reason bit reversal is used in algorithms like Fast Fourier Transform (FFT)?

A To compress data by removing redundant bits

B To encrypt data by scrambling bits

C To convert numbers from decimal to binary faster

D To reorder input data for efficient divide-and-conquer processing

💡 Hint

Think about how FFT processes data in stages.

Bit reversal reorders input indices to enable efficient recursive processing in FFT algorithms.