How to tell the compiler my friend function is function template

Question

Here is my code:

#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
}

I just thought a simple <> notation tell the compiler my function is function template. But I get error with two line:friends can only be classes or functions

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

How can I do to solve it.

Answer 1

Just use this syntax:

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>&);

You nees the first template parameter to be different that T and the second to be different than Sequence, or you will be shadowing the class' template parameters.