如何检查一个类是否在C ++ 11中指定了嵌套类定义或typedef？ [英] How to check whether a class has specified nested class definition or typedef in C++ 11?

问题描述

在我的项目中，我想实现一些现有的更大类的模板代理类。现有的类是库类，因此无法修改。在大多数情况下，客户端不知道对象是代理类还是更大类的实例。但是，在某些情况下，客户必须知道详细的班级信息。由于代理类本身是模板类，因此我认为按类名重载简单函数不会解决此问题。我认为可能的解决方案是在代理类内部添加一个内部嵌套类或typedef，然后客户端检查是否存在该类/ typedef以获取类信息。我的问题是：如何检查类在C ++ 11中是否指定了嵌套类定义或typedef？

In my project, I want to implement a template proxy class of some existing bigger classes. The existing classes are library classes, so they can not be modified. In most cases, the clients do not know the objects are instances of proxy class or bigger class. In some cases, however, the clients MUST know the detailed class information. Since the proxy class is itself a template class, I do not think simple function overloading by class name could solve this problem. The possible solution I thought is to add an internal nested class or typedef inside the proxy class, and the client check whether this class/typedef exists to get the class information. My question is: how to check whether a class has specified nested class definition or typedef in C++ 11 ?

以下代码显示了一个示例：

The following codes show an example:

#include <iostream>
#include <functional>
#include <string>
#include <vector>
#include <type_traits>

typedef std::string CBig1;  //  use string for demonstration
typedef std::string CBig2;  //  use string for demonstration

//class CBig1;   // the bigger class 1, codes of which can not be changed
//class CBig2;   // the bigger class 2, codes of which can not be changed

template <typename _Big, typename _Other>
class CProxy
{
public:
    struct proxy_tag { };
};

//  how to implement this ?
//  the proxy traits class, if defined _T::proxy_tag, the ``type'' will be std::true_type, otherwise the ``type'' will be std::false_type
template <typename _T>
struct is_proxy
{
    //typedef std::true_type type;
    //typedef std::false_type type;
};

template <typename _T>
void ClientHelp(const _T& t, std::false_type)
{
    //  process real class
    std::cerr << "real class" << std::endl;
}

template <typename _T>
void ClientHelp(const _T& t, std::true_type)
{
    //  process proxy class
    std::cerr << "proxy class" << std::endl;
}

template <typename _T>
void Client(const _T& t)
{
    ClientHelp(t, typename is_proxy<_T>::type());
}

int main(int argc, char* argv[])
{
    CBig1 b;
    CProxy<CBig1, int> p;
    Client(b);
    Client(p);
    return 0;
}

如何实现特征类 is_proxy ？

推荐答案

作为C ++ 03版本的补充，在C ++ 11中，您获得 decltype ：

As a complement to the C++03 version, in C++11 you get decltype:

template <typename T>
auto is_proxy(T const&) -> decltype(T::proxy_tag{}, std::true_type{}) {
  return std::true_type{};
}

std::false_type is_proxy(...) { return std::false_type{}; }

您对 Client 的实现变为：

template <typename T>
void Client(T const& t) {
  ClientHelp(t, is_proxy(t));
}

甜，不是吗？

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