未调用移动构造函数

有这么一堂课

 class A
 {
 public:
    A(): _size(0), _value(nullptr)
    {
        std::cout << "A() "<< this << std::endl;
    }
    A(const int size) : _size( size ), _value( new char[size] )
    {
        std::cout << "A( const int ) " << this << std::endl;
        for (int i = 0; i < _size; ++i)
            _value[i] = 'a' + i;
    }

    ~A()
    {
        std::cout << "~A() " << this << std::endl;
        if (_value)
           delete[] _value;
    }

    A( A&& a ):_size(0), _value(nullptr)
    {
        std::cout << "A( A&& ) "<< this << std::endl;
    
        std::swap( _size, a._size );
        std::swap( _value, a._value );
    }

    A& operator=( A&& a )
    {
         std::cout << "operator=( A&& ) "<< this << std::endl;

        std::swap( _size, a._size );
        std::swap( _value, a._value );
     
        return *this;
    }

    A( const A& a)
    {
         std::cout << "A( A& ) "<< this << std::endl;

        _size = a._size;
        _value = new char[_size];

        memcpy( _value, a._value, sizeof( char ) * _size );
    }
    A& operator=( const A& a)
    {
         std::cout << "operator=( const A& ) " << this << std::endl;
    
        _size = a._size;

        if (_value)
            delete _value;

        _value = new char[_size];

        memcpy( _value, a._value, sizeof( char ) * _size );
        return *this;
    }

 private:
    int _size; 
    char* _value;
 };

还有几个功能

A f1(int size)
{
    return A(size);
}

A foo( A&& a )
{
    return A( a );
}

执行此代码时：

 int main()
 {
      A a2 = foo( f1( 5 ) );
      return 0;
 }

输出如下：

 A( const int ) 0050F6E4
 A( A& ) 0050F7B4
 ~A() 0050F6E4
 ~A() 0050F7B4

也就是说，不调用移动构造函数。问题是为什么以及如何让它被调用？