Why is this copy constructor called rather than the move constructor?

问题

The following code snippet causes the copy constructor to be called where I expected the move constructor to be called:

#include <cstdio>

struct Foo
{
    Foo() { puts("Foo gets built!"); }
    Foo(const Foo& foo) { puts("Foo gets copied!"); }
    Foo(Foo&& foo) { puts("Foo gets moved!"); }
};

struct Bar { Foo foo; };
Bar Meow() { Bar bar; return bar; }
int main() { Bar bar(Meow()); }

On VS11 Beta, in debug mode, this prints:

Foo gets built!
Foo gets copied!
Foo gets copied!

I checked the standard and Bar seems to meet all requirements to have a default move constructor automatically generated, yet that doesn't seem to happen unless there's another reason why the object cannot be moved. I've seen a lot of move and copy constructor related questions around here but I don't think anyone has had this specific issue.

Any pointers on what's going on here? Is this standard behaviour?

回答1:

Unfortunately, VS11 doesn't provide a default move constructor. See Move Semantics in the Remarks section - to quote:

Unlike the default copy constructor, the compiler does not provide a default move constructor.

来源：https://stackoverflow.com/questions/10201659/why-is-this-copy-constructor-called-rather-than-the-move-constructor