Return value optimization of tuple/tie

I am looking into return value optimization in the case of tuple/ties and the behavior I observe is not as I expected. In the example below I would expect move semantics to kick in, which it does, but there is one copy operation which remains. The output from the below in optimized is:

Test duo output, non_reference tuple
Default constructor invoked
Parameter constructor invoked
Copy constructor invoked
Move Assignment operator invoked
100

The invocation of the copy constructor in making the tuple inside the function seems unnecessary. Is there any way to remove this? I am using the MSVC 2012 compiler.

#include <iostream>
#include <tuple>

class A
{
public:
     int value;
     A() : value(-1)
     {
         std::cout << "Default constructor invoked" << std::endl;
     }

     explicit A(const int v) : value(v)
     {
         std::cout << "Parameter constructor invoked" << std::endl;
     }

     A(const A& rhs)
     {
         value = rhs.value;
         std::cout << "Copy constructor invoked" << std::endl;
     }

     A(const A&& rhs)
     {
         value = rhs.value;
         std::cout << "Move constructor invoked" << std::endl;
     }

     A& operator=(const A& rhs)
     {
         value = rhs.value;
         std::cout << "Assignment operator invoked" << std::endl;
         return *this;
     }

     A& operator=(const A&& rhs)
     {
         value = rhs.value;
         std::cout << "Move Assignment operator invoked" << std::endl;
         return *this;
     }
 };

 std::tuple<A, int> return_two_non_reference_tuple()
 {
     A tmp(100);

     return std::make_tuple(tmp, 99);
 }

 int main(int argc, char* argv[])
 {

      std::cout << "Test duo output, non_reference tuple" << std::endl;    
      A t3;
      int v1;
      std::tie(t3, v1) = return_two_non_reference_tuple();
      std::cout << t3.value << std::endl << std::endl;

      system("pause");
      return 0;
}

The move constructor will not be called automatically because you are calling

std::make_tuple(tmp, 99);

In this case, tmp is an lvalue. You can use std::move to cast it to an rvalue reference:

return std::make_tuple(std::move(tmp), 99);

This will instruct the compiler to use the move constructor.