Rvalue constructor not being called?

Question

I have this struct

struct A {
    A()             { cout << "Default\n"; }
    A(const A& a)   { cout << "Const lvalue\n"; }
    A(const A&& a)  { cout << "Const rvalue\n"; }
    A(A& a)         { cout << "Lvalue\n"; }
    A(A&& a)        { cout << "Rvalue\n"; }
};

which I'm using to understand rvalues/lvalues.

When doing

A a1;
A a2{ a1 };

const A a3;
A a4{ a3 };

It correctly outputs

> Default
> lvalue
> Default
> Const lvalue

The problem is that when I do something like

A a{ A() };

The output is Default. But isn't A() an rvalue there? Shouldn't A::A(A&&) or A::A(const A&&) have to be called? what is happening here?

Allow me to direct you to [`std::move`](http://en.cppreference.com/w/cpp/utility/move) — , May 08 '17 at 13:58
Thanks, that solved the problem, but why in the example I gave neither any constructor was called? — Garmekain, May 08 '17 at 14:04
Copy constructor is not certain to be called. In this case probably compiler created `A()` directly in place of `a` without copying. — zoska, May 08 '17 at 14:06

score 4 · Accepted Answer · edited May 23 '17 at 12:34

4

You are not seeing Rvalue printed because the compiler is eliding the move. This is allowed even if the constructor has side effects.

If you pass -fno-elide-constructors to GCC or Clang then you will see the output:

clang++ -std=c++14 -O2 -pedantic -fno-elide-constructors main.cpp && ./a.out
Default
Rvalue

clang++ -std=c++14 -O2 -pedantic main.cpp && ./a.out
Default

edited May 23 '17 at 12:34

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answered May 08 '17 at 14:12

TartanLlama

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Rvalue constructor not being called?

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