友元声明声明了一个非模板函数 [英] friend declaration declares a non-template function

问题描述

我有一个类似于下面代码的基类.我试图重载 <<与 cout 一起使用.但是，g++ 说:

I have a base Class akin to the code below. I'm attempting to overload << to use with cout. However, g++ is saying:

base.h:24: warning: friend declaration ‘std::ostream& operator<<(std::ostream&, Base<T>*)’ declares a non-template function
base.h:24: warning: (if this is not what you intended, make sure the function template has already been declared and add <> after the function name here) -Wno-non-template-friend disables this warning

我试过在 << 之后添加 <>在类声明/原型中.但是，然后我得到它不匹配任何模板声明.我一直在尝试将运算符定义完全模板化(我想要)，但我只能使用以下代码使其与手动实例化运算符一起工作.

I've tried adding <> after << in the class declaration / prototype. However, then I get it does not match any template declaration. I've been attempting to have the operator definition fully templated (which I want), but I've only been able to get it to work with the following code, with the operator manually instantiated.

base.h

template <typename T>
class Base {
  public:
    friend ostream& operator << (ostream &out, Base<T> *e);
};

base.cpp

ostream& operator<< (ostream &out, Base<int> *e) {
    out << e->data;
return out;
}

我只想在标题 base.h 中包含此或类似内容:

I want to just have this or similar in the header, base.h:

template <typename T>
class Base {
  public:
    friend ostream& operator << (ostream &out, Base<T> *e);
};

template <typename T>
ostream& operator<< (ostream &out, Base<T> *e) {
    out << e->data;
return out;
}

我在网上的其他地方读到过将 <> 放在 << 之间.原型中的和 () 应该解决这个问题，但事实并非如此.我可以将其放入单个函数模板中吗?

I've read elsewhere online that putting <> between << and () in the prototype should fix this, but it doesn't. Can I get this into a single function template?

推荐答案

听起来你想改变:

friend ostream& operator << (ostream& out, const Base<T>& e);

致:

template<class T>
friend ostream& operator << (ostream& out, const Base<T>& e);

这篇关于友元声明声明了一个非模板函数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！