To start, let’s assume we have a std::bitset named myBits with a size of N. The first step is to convert the std::bitset to a std::vector<int> so that we can sort the individual bits. Here’s how you can do it:

std::vector<int> bitVector;
for (int i = 0; i < N; i++) {
    bitVector.push_back(myBits[i]);
}

Now that we have a std::vector<int> representation of the bits, we can simply use the std::sort algorithm to sort them in ascending order:

std::sort(bitVector.begin(), bitVector.end());

After the sorting is complete, we can copy the sorted bits back to the original std::bitset:

for (int i = 0; i < N; i++) {
    myBits[i] = bitVector[i];
}

And that’s it! You now have a sorted std::bitset in ascending order.

Here’s the complete code snippet for reference:

#include <iostream>
#include <bitset>
#include <vector>
#include <algorithm>

int main() {
    const int N = 8;
    std::bitset<N> myBits("10101010");
    
    std::vector<int> bitVector;
    for (int i = 0; i < N; i++) {
        bitVector.push_back(myBits[i]);
    }
    
    std::sort(bitVector.begin(), bitVector.end());
    
    for (int i = 0; i < N; i++) {
        myBits[i] = bitVector[i];
    }
    
    std::cout << myBits << std::endl;
    
    return 0;
}

In conclusion, while there is no direct method for sorting a std::bitset in C++, you can easily convert it to a std::vector<int>, sort the individual bits, and then copy them back to the original std::bitset. Keep in mind that this sorting method is applicable to small-size bitsets, where the overhead of the conversion and sorting process is manageable.