Let’s consider a scenario where we have a bitset containing a binary representation of a number. Our goal is to count the number of set bits (1s) in the odd and even positions separately.

To begin, let’s assume that our bitset has a fixed size of 8 bits. We can define the bitset and initialize it with a binary number as follows:

#include <bitset>
#include <iostream>

int main() {
    std::bitset<8> bits("10101010");
    std::cout << "Original bitset: " << bits << std::endl;

    return 0;
}

In the above code, we have defined a bitset object called bits with a size of 8 bits. We initialized it with the binary number “10101010”. We have also included the necessary headers for using bitset and iostream in C++.

To count the number of set bits in the odd positions, we can use bit shifting and bitwise AND operations. Here’s the code to accomplish that:

#include <bitset>
#include <iostream>

int countSetBitsOdd(std::bitset<8> bits) {
    // Shift the bits right by 1 position
    bits >>= 1;

    // Perform bitwise AND with a mask to count set bits in odd positions
    bits &= std::bitset<8>("01010101");

    // Count the number of set bits
    return bits.count();
}

int main() {
    std::bitset<8> bits("10101010");
    std::cout << "Original bitset: " << bits << std::endl;

    int setBitsOdd = countSetBitsOdd(bits);
    std::cout << "Number of set bits in odd positions: " << setBitsOdd << std::endl;

    return 0;
}

In the countSetBitsOdd function, we first shift the bits right by 1 position using the >>= operator. This effectively moves all the bits one position to the right. Then, we perform a bitwise AND operation with a mask 01010101, which isolates the bits in odd positions. Finally, we use the count function of the bitset class to count the number of set bits.

Similarly, we can count the number of set bits in the even positions using the following code:

#include <bitset>
#include <iostream>

int countSetBitsEven(std::bitset<8> bits) {
    // Perform bitwise AND with a mask to count set bits in even positions
    bits &= std::bitset<8>("10101010");

    // Count the number of set bits
    return bits.count();
}

int main() {
    std::bitset<8> bits("10101010");
    std::cout << "Original bitset: " << bits << std::endl;

    int setBitsEven = countSetBitsEven(bits);
    std::cout << "Number of set bits in even positions: " << setBitsEven << std::endl;

    return 0;
}

In the countSetBitsEven function, we directly perform a bitwise AND operation with a mask 10101010, isolating the bits in even positions.

By using the above code snippets, you will be able to count the number of set bits in both odd and even positions of a C++ bitset.

#C++ #BitManipulation