如何告诉编译器我的朋友函数是函数模板

Posted 2023-02-22

【中文标题】如何告诉编译器我的朋友函数是函数模板

【英文标题】：How to tell the compiler my friend function is function template

【发布时间】：2016-04-24 08:42:04

【问题描述】：

这是我的代码：

#include <iostream>
#include <cstddef>

class alloc 


;

template <class T, class Alloc = alloc, size_t BufSiz = 0>
class deque

public:
    deque()  std::cout << "deque" << std::endl; 
;

template <class T, class Sequence = deque<T> >
class stack

public:
    stack()  std::cout << "stack" << std::endl; 
private:
    Sequence c;
    friend bool operator== <> (const stack<T, Sequence>&, const stack<T, Sequence> &);
    friend bool operator< <> (const stack<T, Sequence>&, const stack<T, Sequence>&);
;

template <class T, class Sequence>
bool operator== (const stack<T, Sequence>&x, const stack<T, Sequence>&y)

    return std::cout << "opertor == " << std::endl;


template <class T, class Sequence>
bool operator < (const stack<T, Sequence> &x, const stack<T, Sequence> &y)

    return std::cout << "operator <" << std::endl;


int main()

    stack<int> x; // deque stack
    stack<int> y; // deque stack

    std::cout << (x == y) << std::endl; // operator == 1
    std::cout << (x < y) << std::endl; // operator < 1

我只是想一个简单的 符号告诉编译器我的函数是函数模板。但我得到两行错误：朋友只能是类或函数

friend bool operator== <> (const stack<T, Sequence>&, const stack<T, Sequence> &);
friend bool operator< <> (const stack<T, Sequence>&, const stack<T, Sequence>&);

我该怎么做才能解决它。

【参考方案1】：

只需使用以下语法：

template<typename T1, typename Sequence1>
friend bool operator== (const stack<T1, Sequence1>&, const stack<T1, Sequence> &);

template<typename T1, typename Sequence1>
friend bool operator< (const stack<T1, Sequence1>&, const stack<T1, Sequence>&);

您需要第一个模板参数与 T 不同，第二个模板参数与 Sequence 不同，否则您将隐藏类的模板参数。