Why is the move constructor not called?

Question

As per my understanding, the move constructor will be called when there is a temporary object created. Here the getA() function is returning a temporary object but my program is not printing the message from the move constructor:

#include <iostream>

using namespace std;

class A
{
    public:
    A()
    {
        cout<<"Hi from default\n";
    }

    A(A && obj)
    {
        cout<<"Hi from move\n";
    } 
};

A getA()
{
    A obj;
    cout<<"from getA\n";
    return obj;
}

int main()
{
    A b(getA());

   return 0;
}

I was searching with move constructor and google is not leading me to that page — Gilson PJ, Jul 07 '16 at 14:56

Bathsheba · Accepted Answer · 2016-07-07T13:46:19.817

9

The compiler is allowed to optimise out the instance obj and send the object directly back to the caller without a conceptual value copy being taken.

This is called named return value optimisation (NRVO). It's a more aggressive optimisation than classical return value optimisation (RVO) that a compiler can invoke to obviate the value copy of an anonymous temporary.

For the avoidance of doubt the compiler can do this even if there is a side-effect in doing so (in your case the lack of console output).

edited Jul 07 '16 at 13:46

answered Jul 07 '16 at 13:33

Bathsheba

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What was wrong with the dupe target? – NathanOliver Jul 07 '16 at 14:13
It was more to do with RVO not (the more recent) NRVO. IMHO a duplicate is only a good one if it's pretty exact. – Bathsheba Jul 07 '16 at 14:17
Half of the answers talked about and example NRVO. – NathanOliver Jul 07 '16 at 14:20

Why is the move constructor not called?

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