Is this normal behavior for std::bitset::operator^= and std::bitset::count ? If so, why? [closed]

Question

As documented here, std::bitset::operator^= returns *this. From that and from the "usual" interpretation of operators such as +=, |=, *= one could reasonably assume that given std::bitset instances (of the same size) a and b, the expression (a^=b).count() will store the result of a bitwise XOR operation in a, and that count() would return the number of bits in a that are set to true. However, as the following minimal example demonstrates, something unexpected happens:

#include <iostream>
#include <bitset>

int main()
{
    constexpr unsigned int N=6; 
    std::bitset<N> a; 
    std::bitset<N> b; 

    a.flip();//111111
    b[0]=1;
    b[4]=1;//b is now 010001 (assuming least significan bit on the right end of the string)

    std::cout<<"a=="<<a.to_string()<<std::endl;
    std::cout<<"b=="<<b.to_string()<<std::endl;
    std::cout<<"(a xor b) to string=="<<(a^=b).to_string()<<std::endl;

    //Here is the unexpected part!
    std::cout<<"(a xor b) count=="<<(a^=b).count()<<std::endl;
    //Note that the following lines would produce the correct result
    //a^=b;
    //std::cout<<a.count()<<std::endl;


    return 0;
}

The output is

a==111111
b==010001
(a xor b) to string==101110       
(a xor b) count==6                //this is wrong!!!!! It should be 4...

A quick look at the implementation of std::bitset (see here) seems to indicate that the reference that is returned is indeed a reference to the lhs object (a in my example). So... Why is this happening?

Answer 1

This has nothing to do with the bitset. Consider this code:

int a = 2;
int b = 3;
std::cout << std::to_string(a *= b) << std::endl; // Prints 6.
std::cout << std::to_string(a *= b) << std::endl; // Prints 18.

You are using an assignment operator, so your variable/bitset changes every time. In your case, the second evaluation yields ((a ^ b) ^ b), which is of course the original a (which did have 6 bits set).