How does copy constructor that returns value, discards the temp?-CodePudding

having this code:

#include <iostream>

class Base {
public:
    Base() = default;

    explicit Base(int val) : _var(val) {}

    Base operator=(const Base &rhs) {
        _var = rhs._var;
        return *this;
    }

    void print() const {
        std::cout << _var << std::endl;
    }

private:
    int _var;
};

int main() {
    Base b[] = {Base(10), Base(), Base(), Base()};
    (b[1] = b[2]) = b[0];
    for (Base base: b) {
        base.print();
    }
}

the output is:

but I would expect

As the second element in array b[1] should get assign from b[0], but the assignment operator returns value, not reference and thus copy-constructing happen. But still, why is not b[1] copy-constructed to have _var=10?

If the operator= returned Base &, the output would be my expectation

CodePudding user response：

To get the desired result of your assignment operator (which, by the way, is different from copy constructor), you need to return a reference:

Base& operator=(const Base &rhs)

This is the canonical form.

Without the reference, the result of (b[1] = b[2]) is stored in a temporary. (b[1] = b[2]) = b[0]; assigns to that temporary, which is discarded and has no effect on b[1].