为什么 C++11 的移动构造函数/赋值运算符没有按预期运行

#include <iostream>

using namespace std;

struct A
{
    A()
    {
        cout << "A()" << endl;
    }

    ~A()
    {
        cout << "~A()" << endl;
    }

    A(A&&)
    {
        cout << "A(A&&)" << endl;
    }

    A& operator =(A&&)
    {
        cout << "A& operator =(A&&)" << endl;
        return *this;
    }
};

struct B
{
    // According to the C++11, the move ctor/assignment operator
    // should be implicitly declared and defined. The move ctor
    // /assignment operator should implicitly call class A's move
    // ctor/assignment operator to move member a.
    A a;
};

B f()
{
    B b;

    // The compiler knows b is a temporary object, so implicitly 
    // defined move ctor/assignment operator of class B should be
    // called here. Which will cause A's move ctor is called.
    return b; 
}

int main()
{
    f();
    return 0;
}

我的预期输出应该是：

A()
A(A&&)
~A()
~A()

然而，实际输出是：（C++编译器是：Visual Studio 2012）

A()
~A()
~A()

这是VC++的bug吗？还是只是我的误会？

根据这篇博文 http://blogs.msdn.com/b/vcblog/archive/2011/09/12/10209291.aspx，VC++ 2012目前实现N2844 http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2009/n2844.html + DR1138 http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_defects.html#1138， 但不是N3053 http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2010/n3053.html。结果，编译器是not为您隐式生成移动构造函数或赋值运算符。如果添加显式默认值并将构造函数移动到B然后你会得到你期望的输出。