Question

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＆lt;＆gt;符号告诉编译器我的函数是函数模板。但我得到两行错误：朋友只能是班级或职能

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。

Answer 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>&);

第一个模板参数与T不同，第二个模板参数与Sequence不同，或者您将隐藏类的模板参数。

如何告诉编译器我的朋友功能是功能模板

1 个答案: