在成员函数中使用两个模板的类模板中为成员函数定义单个模板 [英] Defining a Single Template for a Member Function within a Class Template with Both Templates Used in Member Function

问题描述

我目前正在学习模板在 C++ 中的工作原理.特别是，我正在查看类模板中的单个成员函数模板.要理解我的意思，请在下面找到代码.

I'm currently learning how templates work in C++. In particular, I'm looking at the single member function templates within class templates. To understand what I mean, code is found below.

// foo.h
template<typename A>
class foo {
    template<typename B>
    void boo(B);
};

// foo.cpp
template<typename A>
void foo<A>::boo(B value) {} // compiler error: 'Unknown' type name B

// or if I try this

template<typename B>
void foo<A>::boo(B value) {} // compiler error: Use of undeclared identifier A

我正在尝试使用两个类型名，一个来自类模板，一个来自单个文件模板，用于该特定功能.但是在上面的两个版本中，我遇到了编译器错误.我该如何绕过这个问题?

I'm trying to use two typenames, one from the class template, and one from the single file template, for that specific function. But in those two versions above, I get compiler errors. How would I bypass this problem?

推荐答案

这两组模板参数都需要定义成员模板.

Both the two sets of template parameters are required to define the member template.

(强调我的)

如果封闭类声明又是一个类模板，当在类主体之外定义成员模板时，它需要两组模板参数:一组用于封闭类，另一组用于本身:

If the enclosing class declaration is, in turn, a class template, when a member template is defined outside of the class body, it takes two sets of template parameters: one for the enclosing class, and another one for itself:

例如

template<typename A>
template<typename B>
void foo<A>::boo(B value) {} 

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