左值引用类型成员的用户定义移动构造函数

Question

我正在研究编译器上的移动语义，它具有右值引用但不支持默认移动构造函数。我想生成类似下面的包装器 class 的东西，即使模板参数是左值引用，它也能工作。但是，这种直接的方法无法编译，因为它试图从 int 初始化 int&。

#define USER_DEFINED   0

template <typename T>
struct Wrapper
{
    Wrapper(T t)
        : m_t(t)
    {
    }

    Wrapper(const Wrapper&) = delete;
    Wrapper& operator=(const Wrapper&) = delete;

#if USER_DEFINED
    Wrapper(Wrapper&& w)
        : m_t(std::move(w.m_t))
    {
    }
#else
    Wrapper(Wrapper&&) = default;
#endif

private:
    T m_t;
};

int main()
{
    int i = 0;
    Wrapper<int&> w1 = Wrapper<int&>(i);
    Wrapper<std::string> w2 = Wrapper<std::string>("text");
}

显而易见的解决方案是拥有两个移动构造函数，一个用于左值引用，一个用于所有其他类型。例如：

template <typename U = T>
Wrapper(typename std::enable_if<!std::is_lvalue_reference<U>::value, Wrapper>::type&& w)
    : m_t(std::move(w.m_t))
{
}

template <typename U = T>
Wrapper(typename std::enable_if<std::is_lvalue_reference<U>::value, Wrapper>::type&& w)
    : m_t(w.m_t)
{
}

这是要走的路吗？也许 enable_if<> 中的表达式应该更通用？或者我可以使用不同于 std::move() 的东西，并为所有类型使用一个构造函数吗？

Answer 1

好的，这里有一个解决方案，我认为它可以按照您的意愿工作，但我必须承认我并不完全理解它是如何工作的。

我所做的最重要的更改是将 std::move 替换为 std::forward<T> in the move constructor. I have also added a move assignment operator but here is the thing I don't understand: Unless everywhere else, it needs std::move instead of std::forward! Finally, I have also added a std::forward to your constructor that accepts a T so it doesn't make two copies of its argument. It is actually required that we accept the T by value and use std::forward here. Overloading for const T& and T&& would fail if T is a reference because then T&& would also collapse 到左值引用，并且重载变得不明确。

#include <iostream>
#include <utility>

template <typename T>
class Wrapper
{

public:

  Wrapper(T t) : m_t {std::forward<T>(t)}
  {
  }

  Wrapper(Wrapper&& w) : m_t {std::forward<T>(w.m_t)}
  {
  }

  Wrapper(const Wrapper&) = delete;

  Wrapper&
  operator=(Wrapper&& w)
  {
    if (this != &w)
      this->m_t = std::move(w.m_t);
    return *this;
  }

  Wrapper&
  operator=(const Wrapper&) = delete;

private:

  T m_t;
};

现在，让我们用仪表类型试驾一下，让我们看看发生了什么。

struct A
{
  A ()
  {
    std::cerr << "A was default-constructed" << std::endl;
  }

  A (const A&)
  {
    std::cerr << "A was copy-constructed" << std::endl;
  }

  A (A&&)
  {
    std::cerr << "A was move-constructed" << std::endl;
  }

  A&
  operator=(const A&)
  {
    std::cerr << "A was copy-assigned" << std::endl;
    return *this;
  }

  A&
  operator=(A&&)
  {
    std::cerr << "A was move-assigned" << std::endl;
    return *this;
  }

  ~A ()
  {
    std::cerr << "A was destroyed" << std::endl;
  }
};

int main()
{
  A a {};
  Wrapper<A> val1 {a};
  Wrapper<A> val2 {std::move(val1)};
  val1 = std::move(val2);
  Wrapper<A&> ref1 {a};
  Wrapper<A&> ref2 {std::move(ref1)};
  ref1 = std::move(ref2);
}

使用 GCC 4.9.1 编译（整个练习实际上毫无意义，因为它支持开箱即用的各种移动。）并关闭所有优化，输出如下（已添加注释）。

A was default-constructed  ; automatic variable a in main
A was copy-constructed     ; copied as Wrapper<A>'s constructor argument
A was move-constructed     ; forwarded in Wrapper<A>'s initializer
A was destroyed            ; the moved-away from copy as the constructor returns
A was move-constructed     ; forwarded in Wrapper<A>'s move-constructor
A was move-assigned        ; move-assignment operator in Wrapper<A>
A was move-assigned        ; not sure about this one... (apparently caused by the Wrapper<A&> move-assignment)
A was destroyed            ; the sub-object of val2
A was destroyed            ; the sub-object of val1
A was destroyed            ; the automatic variable a in main

左值引用类型成员的用户定义移动构造函数

User-defined move constructor for member of lvalue reference type

c++

move-semantics

c++11