通过基类专门化类模板 [英] Specializing a class template by a base class

问题描述

我对以下这段代码提出疑问：

I have distilled my doubt to this following piece of code

struct base {};
struct derived : public base {};

template <class T>
struct Type { };

template <> struct Type<base> {
  typedef float mytype;
};

typename Type<base>::mytype a=4.2;    // this works
typename Type<derived>::mytype a=4.2; // this doesnt

有人可以解释为什么我不能用 derived 并建议一个简单的方法来做。对于我感兴趣的实际问题有许多派生类使用它我想要模板类对象和/或使用typedefs。

Could anyone explain why I cannot intantiate the class template object with derived and suggest a simple way to do it. For the actual problem that I am interested in there are many derived classes using which I want to intantiate template class objects and/or use typedefs. There are too many of them than what I would want to specialize individually.

编辑：忘记了，我的坏，这需要是C ++ 03

Forgot to mention, my bad, this needs to be C++03

推荐答案

#include <iostream>
#include <type_traits>

struct base { };
struct derived : base { };

template<typename T, bool = std::is_base_of<base, T>::value>
struct Type { };

template<typename T>
struct Type<T, true>
{
   typedef float mytype;
};

int main()
{
   Type<base>::mytype a1 = 4.2f;
   Type<derived>::mytype a2 = 8.4f;
   std::cout << a1 << '\n' << a2 << '\n';
}

在C ++ 03中， std 可以简单地替换为 boost ： boost :: is_base_of

In C++03, std can be trivially replaced with boost: boost::is_base_of

这篇关于通过基类专门化类模板的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！