CODEWITHSUNDEEP
c++templatesfriend-function
joke
joke

Reputation: 658

How to split the definition of template friend function within template class?

The following example compiles fine but I can't figure out how to separate declaration and definition of operator<<() is this particular case.

Every time I try to split the definition friend is causing trouble and gcc complains the operator<<() definition must take exactly one argument.

#include <iostream>
template <typename T>
class Test {
    public:
        Test(const T& value) : value_(value) {}

        template <typename STREAM>
        friend STREAM& operator<<(STREAM& os, const Test<T>& rhs) {
            os << rhs.value_;
            return os;
        }
    private:
        T value_;
};

int main() {
    std::cout << Test<int>(5) << std::endl;
}

Operator<<() is supposed to have a free first parameter to work with different kind of output streams (std::cout, std::wcout or boost::asio::ip::tcp::iostream). The second parameter should be bound to a specialized version of the surrounding class.

Test<int> x;
some_other_class y;

std::cout << x; // works
boost::asio::ip::tcp::iostream << x; // works

std::cout << y; // doesn't work
boost::asio::ip::tcp::iostream << y; // works

Besides that using a non-member-function isn't equivalent to splitting the definition and declaration because non-member-functions can't access private attributes the the class.

Upvotes: 5

Views: 1049

Answers (5)

joke
joke

Reputation: 658

For each instantiated type T of class Test a template function operator<<() is exposed which can operate on different kind of streams. The operator<<() function has a free first parameter but a fixed second parameter.

example:

Test<int> x;
some_other_class y;

std::cout << x; // works
boost::asio::ip::tcp::iostream << x; // works

std::cout << y; // doesn't work
boost::asio::ip::tcp::iostream << y; // works

That's the way the Test class was supposed to work.

Upvotes: 0

David Rodr&#237;guez - dribeas
David Rodr&#237;guez - dribeas

Reputation: 208353

The problem is that in the code that you present the friend is a templated function parametrized only on the first argument type. That is, for each instantiating type T of the class template (call it mytype), you are declaring a free template function:

template <typename STREAM>
STREAM& operator<<( STREAM& os, Test<mytype> const & x );

The important point there is that Test<mytype> is the particular instantiation of Test with type argument mytype.

If you really want to declare a friend function that is templated in both the stream type and the instantiating type of the Test template, you must declare a friend with two arguments.

On the other hand, I recommend that you do not parametrize operator<< on the stream type, and at the same time, that you define it inside the class braces as it has slight advantages (name lookup rules are slightly different).

Upvotes: 0

a1ex07
a1ex07

Reputation: 37364

Shouldn't it be defined outside of the class ? 

template <typename T>
class Test 
{  
    ...
    template <typename STREAM>
    friend STREAM& operator<<(STREAM& os, const Test<T>& rhs);
};

template <typename STREAM, typename T> 
STREAM& operator<<(STREAM& os, const Test<T>& rhs) 
{
    os << rhs.value_;
    return os;
}

Upvotes: 1

AProgrammer
AProgrammer

Reputation: 52294

The nearest I can achieve is

#include <iostream>

template <typename T>
class Test;

template <typename STREAM, typename T>
STREAM& operator<<(STREAM& os, const Test<T>& rhs);

template <typename T>
class Test {
public:
    Test(const T& value) : value_(value) {}

    template <typename STREAM, typename U>
    friend STREAM& operator<< (STREAM& os, const Test<U>& rhs);

private:
    T value_;
};

template <typename STREAM, typename T>
STREAM& operator<<(STREAM& os, const Test<T>& rhs) {
    os << rhs.value_;
    return os;
}

int main() {
    std::cout << Test<int>(5) << std::endl;
}

which declares all operator<< as friend instead of only the one parametrized by T. The problem is that it isn't possible to partially specialize functions. One would have liked to use

template <typename STREAM>
friend STREAM& operator<< <STREAM, T> (STREAM& os, const Test<T>& rhs);

but that isn't valid syntax. (Well, and partial specialization can't declared friend)

Upvotes: 1

Nikolai Fetissov
Nikolai Fetissov

Reputation: 84169

The easiest is probably to make all these template operators friends:

#include <iostream>
template <typename T>
class Test
{
    public:
        Test(const T& value) : value_(value) {}

        template <typename STREAM, typename U>
        friend STREAM& operator<<(STREAM& os, const Test<U>& rhs);

    private:
        T value_;
};

template <typename STREAM, typename T>
STREAM& operator<<( STREAM& os, const Test<T>& rhs )
{
    os << rhs.value_;
    return os;
}

Upvotes: 4

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