模板类成员函数只专门化 [英] template class member function only specialization

问题描述

我正在阅读模板的完整指南，它说明了以下内容：

I am reading the Complete Guide on Templates and it says the following:

它正在谈论类模板专业化。

Where it is talking about class template specialization.

虽然
可能专门用于单个成员
函数，但一旦你这样做，你
就不能再专门化整个
类。

However, if you specialize a class template, you must also specialize all member functions. Although it is possible to specialize a single member function, once you have done so, you can no longer specialize the whole class.

我真的想知道这是如何是真的，因为你可以专门化没有任何成员函数。它是说，你不能有一个专门化只有一个成员函数，然后另一个与所有成员函数？

I'm actually wondering how this is true, cause you can specialize without any member functions at all. Is it saying that you cannot have a specialization with only one member function and then another with all member functions?

有人可以澄清吗？

推荐答案

对于以下情况：

template <typename T>
struct base {
   void foo() { std::cout << "generic" << std::endl; }
   void bar() { std::cout << "bar" << std::endl; }
};
template <>
void base<int>::foo() // specialize only one member
{ 
   std::cout << "int" << std::endl; 
}
int main() {
   base<int> i;
   i.foo();         // int
   i.bar();         // bar
}

完成后，您就无法将完整模板任何其他事物，因此

Once that is done, you cannot specialize the full template to be any other thing, so

template <>
struct base<int> {};  // error

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