Why does my rvalue copy constructor not work? [duplicate] - c++

Question

This question already has answers here:
What are copy elision and return value optimization?
(5 answers)
Closed 5 years ago.
I was testing my knowledge in C++ and have encountered a problem. Consider this code:
class A
{
public:
A(int n = 0)
: m_n(n)
{
std::cout << 'd';
}
A(const A& a)
: m_n(a.m_n)
{
std::cout << 'c';
}
A(A&& a){
std::cout<<'r';
}
private:
int m_n;
};
int main()
{
A a = A();
std::cout << std::endl;
return 0;
}
Clearly, the A() constructor is an rvalue, as no permanent object has been created. So I think that first I have to see "d" as output. Then we are copying the rvalue to our new object, which is yet to be initialised. I have implemented a copy constructor that accepts an rvalue as an argument but I did not see the proof (no "r" output).
Can someone explain why that is?

Answer 1

You're seeing a form of copy elision -- the compiler is actually required to optimize and implement A a = A() by initializing a directly with the arguments to the constructor expression (() in this case) rather than creating a temporary and copying it into a. If you want to enforce an unnamed object to be copied, you need to initialize a with an expression that is not a simple constructor call.
Note that even then, the compiler may elide construtor calls by the as-if rule (if the only visible side effects are in the copy/move constructors/destructors. Your best bet to "force" a call to the move ctor is to use a named value and std::move:
A a1;
A a2 = std::move(a1);