是否可以使用成员枚举来专门化模板？ [英] Is it possible to specialize a template using a member enum?

问题描述

struct Bar {
  enum { Special = 4 };
};

template<class T, int K> struct Foo {};
template<class T> struct Foo<T,T::Special> {};

用法：

Foo<Bar> aa;

无法使用gcc编译4.1.2
它抱怨使用 T :: Special 用于部分指定Foo。如果 Special 是一个类，解决方案将是它前面的typename。

fails to compile using gcc 4.1.2 It complains about the usage of T::Special for partial specilization of Foo. If Special was a class the solution would be to a typename in front of it. Is there something equivalent to it for enums (or integers)?

推荐答案

因为这不是C ++允许的推荐 Prasoon，所以一个替代的解决方案是使用 EnumToType 类模板

Since that is not allowed by C++ as explained by Prasoon, so an alternative solution would be to use EnumToType class template,

struct Bar {
  enum { Special = 4 };
};

template<int e>
struct EnumToType
{
  static const int value = e;
};

template<class T, class K> //note I changed from "int K" to "class K"
struct Foo
{};

template<class T> 
struct Foo<T, EnumToType<(int)T::Special> > 
{
   static const int enumValue = T::Special;
};

在ideone的示例代码： http://www.ideone.com/JPvZy

Sample code at ideone : http://www.ideone.com/JPvZy

或者，您可以简单地（如果它解决了您的问题），

Or, you can simply specialize like this (if it solves your problem),

template<class T> struct Foo<T,Bar::Special> {};

//usage
Foo<Bar, Bar::Special> f;

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