时间:2019-03-17 标签:c++: how to remove cv-qualifiers of a type to access class functions?

Question

这里是一个例子:

#include <iostream>

template<typename T, 
         typename ... Args>
void print(T&& t, Args&& ... args)
{
    // line where compilation fails when the A::run is called
    if constexpr (std::is_invocable_v<decltype(&T::display),T*,Args...>)
      {
          t.display(std::forward<Args>(args)...);
      }
    else 
    {
        std::cout << "not applicable !" << std::endl;    
    }
}

template<typename T>
class A 
{
    public:

    A(T&& t):t_(t){}

    template <typename... Args>
    void run(Args&& ... args)
    {
        print<T,Args...>(t_,std::forward<Args>(args)...);
    }

    T t_;
};

template <typename T> A(T&) -> A<T&>;
template <typename T> A(T&&) -> A<T>;

class B 
{
  public:
  B(int value):value_(value){} 
  void display(int a, int b)
  {
      std::cout << value_ << " " 
                << a << " " 
                << b << std::endl; 
  }
  int value_;
};

int main()
{
    int v1=10;
    int v2=20;

    B b1{1};
    A a1{b1};
    a1.t_.display(v1,v2);

    A a2{B(2)};
    a2.t_.display(v1,v2);

    //a1.run(v1,v2); // (1)
    //a2.run(v1,v2); // (2)
    //a1.run(v1);

    return 0;
}

上面的代码编译并运行良好。但如果最后 3 行(调用 run() )未注释，则会出现以下编译错误:

(1)

main.cpp:7:48: error: ‘display’ is not a member of ‘B&’

if constexpr (std::is_invocable_v<decltype(&T::display),T*,Args...>)

(2)

main.cpp:27:25: error: no matching function for call to ‘print(B&, int&, int&)’

    print<T,Args...>(t_,std::forward<Args>(args)...);

注意:

template <typename T> A(T&) -> A<T&>;
template <typename T> A(T&&) -> A<T>;

解释如下:

c++ copy (reference) constructor and move constructor of class cohabitation

Answer 1

问题(1)和(2)是不同的问题。

问题(1)来自以下事实:在以下std::is_invocable_v

template<typename T, 
         typename ... Args>
void print(T&& t, Args&& ... args)
{
    if constexpr (std::is_invocable_v<decltype(&T::display),T*,Args...>)
    //  no T but std::decay_t<T> ...............^...........^

您需要的是“decayed”类型，而不是T(可以作为引用)

我提议

template <typename T, typename ... Args>
void print (T && t, Args && ... args)
 {
   using U = std::decay_t<T>;

   if constexpr ( std::is_invocable_v<decltype(&U::display), U*, Args...> )
      t.display(std::forward<Args>(args)...);
   else 
      std::cout << "not applicable !" << std::endl;    
}

问题 (2) 来自以下事实:在以下 print() 调用中解释模板参数

template <typename... Args>
void run(Args&& ... args)
{
    print<T,Args...>(t_,std::forward<Args>(args)...);
}

你妨碍了正确的模板推导。

建议:让模板推导，完美转发，工作原理如下

    print(t_,std::forward<Args>(args)...);