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:
10
0
0
0
but I would expect
10
10
0
0
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
>Solution :
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].