给定GMan's美味邪恶auto_cast炮制效用函数 here ，我一直在试图弄清楚为什么当我尝试 auto_cast 时它不为我编译来自右值(在 MSVC 10.0 上)。

这是我正在使用的代码:

template <typename T>
class auto_cast_wrapper : boost::noncopyable
{
  public:
    template <typename R>
    friend auto_cast_wrapper<R> auto_cast(R&& pX);

    template <typename U>
    operator U() const
    {
      return static_cast<U>( std::forward<T>(mX) );
    }

  private:
    //error C2440: 'initializing': cannot convert from 'float' to 'float &&'
    auto_cast_wrapper(T&& pX) : mX(pX) { }

    T&& mX;
};

template <typename R>
auto_cast_wrapper<R> auto_cast(R&& pX)
{
  return auto_cast_wrapper<R>( std::forward<R>(pX) );
}

int main()
{
  int c = auto_cast( 5.0f );  // from an rvalue
}

尽我所能，我已尝试遵循 C++0x 引用折叠规则和概述的模板参数推导规则 here ，据我所知，上面给出的代码应该可以工作。

Recall that in pre-0x C++, it is not allowed to take a reference to a reference: something like A& & causes a compile error. C++0x, by contrast, introduces the following reference collapsing rules:

• A& & becomes A&
• A& && becomes A&
• A&& & becomes A&
• A&& && becomes A&&

The second rule is a special template argument deduction rule for function templates that take an argument by rvalue reference to a template argument:

template<typename T>  
void foo(T&&);

Here, the following rules apply:

1. When foo is called on an lvalue of type A, then T resolves to A& and hence, by the reference collapsing rules above, the argument type effectively becomes A&.
2. When foo is called on an rvalue of type A, then T resolves to A, and hence the argument type becomes A&&.

现在，当我将鼠标悬停在对 auto_cast( 5.0f ) 的调用上时, 工具提示正确地将其返回值显示为 auto_cast_wrapper<float> .这意味着编译器已正确遵循规则 2:

When foo is called on an rvalue of type A, then T resolves to A.

所以因为我们有一个 auto_cast_wrapper<float> , 构造函数应该实例化一个 float&& .但是错误消息似乎暗示它实例化了一个float。按值(value)。

这是完整的错误消息，再次显示 T=float 正确但 T&& 参数变为 T？

 main.cpp(17): error C2440: 'initializing' : cannot convert from 'float' to 'float &&'
     You cannot bind an lvalue to an rvalue reference
     main.cpp(17) : while compiling class template member function 'auto_cast_wrapper<T>::auto_cast_wrapper(T &&)'
     with
     [
         T=float
     ]
     main.cpp(33) : see reference to class template instantiation 'auto_cast_wrapper<T>' being compiled
     with
     [
         T=float
     ]

有什么想法吗？

最佳答案

您忘记将 std::forward T&& 参数传递给 auto_cast_wrapper 构造函数。这打破了转发链。编译器现在发出警告，但它似乎工作正常。

template <typename T>
class auto_cast_wrapper
{
  public:
    template <typename R>
    friend auto_cast_wrapper<R> auto_cast(R&& pX);

    template <typename U>
    operator U() const
    {
      return static_cast<U>( std::forward<T>(mX) );
    }

  private:
    //error C2440: 'initializing': cannot convert from 'float' to 'float &&'
    auto_cast_wrapper(T&& pX) : mX(std::forward<T>(pX)) { }

    auto_cast_wrapper(const auto_cast_wrapper&);
    auto_cast_wrapper& operator=(const auto_cast_wrapper&);

    T&& mX;
};

template <typename R>
auto_cast_wrapper<R> auto_cast(R&& pX)
{
  return auto_cast_wrapper<R>( std::forward<R>(pX) );
}

float func() {
    return 5.0f;
}

int main()
{

  int c = auto_cast( func() );  // from an rvalue
  int cvar = auto_cast( 5.0f );

  std::cout << c << "\n" << cvar << "\n";
  std::cin.get();
}

打印一对五。

关于C++0x 右值引用模板参数推导，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/4033964/