Default Move Constructor in Visual Studio 2013 (Update 3)

Question

I've been able to find multiple conversations about this in the past (e.g. here), but such conversations are from quite a while ago. The code I have a question about is:

#include <utility>
#include <iostream>

struct Foo
{
    Foo() = default;
    Foo(const Foo &o)
    {
        std::cout << "copy" << std::endl;
    }
    Foo(Foo &&o)
    {
        std::cout << "move" << std::endl;
    }
};

struct Bar
{
    Foo foo;
};

int main(void)
{
    Bar a;
    Bar b(a);
    Bar c(std::move(a));
}

If you execute the code in Visual Studio 2013 (Update 3), it prints out "copy" for both cases. If the standard hasn't changed since the answer in the link above, then the output should be "copy" followed by "move." Ideone seems to give the correct output. Is this just something that Visual Studio hasn't implemented yet, or is there something missing in my code? I know that you cannot mark move constructors as default, but that does not imply that the compiler does not support generating default move constructors all-together.

Answer 1

I know that you cannot mark move constructors as default, but that does not imply that the compiler does not support generating default move constructors all-together

Unfortunately, that's exactly what that means. VS2013 does not support implicit generation of move constructors and move assignment operators. If it did, they wouldn't really have a reason to disallow the = default syntax, especially since you're allowed to do that for the copy constructor and assignment operator.

Quoting MSDN: Support For C++11 Features (Modern C++)

"Rvalue references v3.0" adds new rules to automatically generate move constructors and move assignment operators under certain conditions. However, this is not implemented in Visual C++ in Visual Studio 2013, due to time and resource constraints.