如何从C ++模板中的方法类型推导类类型？ [英] How to deduce class type from method type in C++ templates?

问题描述

在如下所示的模板中，我想调用 Run（& Base :: foo）成功，而不需要将Base类型命名两次在编译 Run< Base>（& Base :: foo）调用中）。我可以有吗？可能没有添加大量的 Boost 标头？

In templates as shown below, I would like the call Run(&Base::foo) succeed without the need to name the Base type twice (as is done in the compiling Run<Base>(&Base::foo) call). Can I have that? Possibly without adding a ton of Boost headers?

使用提供的代码，我得到一个错误：

With the provided code, I get an error of:

prog.cpp:26: error: no matching function for call to ‘Run(bool (Base::*)())’

（你可以使用 http://ideone.com/8NZkq ）：

#include <iostream>

class Base {
public:
  bool foo() { return true; }
};

Base* x;

template<typename T>
struct Traits {
  typedef bool (T::*BoolMethodPtr)();
};

template<typename T>
void Run(typename Traits<T>::BoolMethodPtr check) {
  T* y = dynamic_cast<T*>(x);
  std::cout << (y->*check)();
}

int main() {
  Base y;
  x = &y;
  Run<Base>(&Base::foo);
  Run(&Base::foo); // why error?
}

推荐答案

中的 T应该。
这是因为可能有这样的代码：

The T in Traits<T>::BoolMethodPtr is in a non-deduced context, so the compiler will not deduce automatically from the call what type T should be. This is because there could be code like this:

template<typename T>
struct Traits {
  typedef bool (T::*BoolMethodPtr)();
};

template<>
struct Traits<int> {
  typedef bool (Base::*BoolMethodPtr)();
};

Run(&Base::foo); /* What should T be deduced to? Base and int are both equally possible */

如果你不需要 Traits< T> 类，您可以写运行为：

If you can do without the Traits<T> class, you can write Run as:

template<class Class>
void Run(bool (Class::*check)()) {
  Class* y = dynamic_cast<Class*>(x);
  std::cout << (y->*check)();
}

在此上下文中， c>可以推导为 Base

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