部分专门化成员函数实现 [英] Partially specializing member-function implementations

问题描述

我目前正在重构一些代码，该代码明确专用于具有两个模板参数的类模板的成员函数.

I'm currently refactoring some code the explicitly specializes a member function of a class template with two template parameters.

template <class S, class T>
class Foo
{
  void bar();
};

template <class S, class T>
void Foo<S, T>::bar()
{ /* Generic stuff */ }

template <>
void Foo<SomeType, SomeType>::bar()
{ /* Some special function */ }

现在，我添加了更多模板参数，因此该类现在看起来像这样:

Now I added some more template parameters, so the class now looks like this:

template <class S, class EXTRA0, class T, class EXTRA1>
class Foo
{
  void bar();
};

这两个额外的参数只是将typedef添加到我的类中，因此运行时功能并没有真正改变.有什么方法可以保留bar的(现在是部分)专业化的实现?我似乎无法弄清楚它的语法，并且我预感这可能是不可能的.

These two extra parameters just add typedefs to my class, so the run-time functionality doesn't really change. Is there any way I can keep the (now partially) specialized implementation of bar? I can't seem to figure out the syntax for that and I have a hunch that it might not be possible.

我正在寻找类似的东西

I'm looking for something like:

template <class EXTRA0, class EXTRA1>
void foo<SomeType, EXTRA0, Sometype, EXTRA1>::bar()
{
   /* specialized implementation */
}

似乎没有编译..

推荐答案

您是正确的，这是不可能的.

You are correct, it is not possible.

您可以做的是在新的 Foo 内创建一个辅助成员类模板，并将专用函数作为非模板成员函数放在其中.专门使用助手类而不是函数.

What you can do is create a helper member class template inside the new Foo, and place the specialized function inside it as a non-template member function. Specialize the helper class instead of the function.

另一种选择是将专业化转化为非模板重载.

Another alternative is to turn the specialization into a non-template overload.

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