Friend function template class

Question

0.00/5 (No votes)

See more:

I know of two ways to deal with friend function of a template class.
The first way is make the friend function inline:
header file A.h

C++

#ifndef A_H
#define A_H
#include<iostream>
using namespace std;
template <class T>
class A
{
	friend ostream& operator << (ostream& outs, const A<T>& object)
	{
		//definition goes here
	}
	//Other functions of class A
}
#endif //A_H

Implementation file A.cpp

C#

//the rest of the function definition of class A

The second way is forward declare class A and the friend function and then you put the friend function definition and other definitions of functions of class A in implementation file
header file A.h

C++

#ifndef A_H
#define A_H
#include<iostream>
using namespace std;
template <class T>
class A;
template <class T>
ostream& operator << (ostream& outs, const A<T>& object);
class A
{
	friend ostream& operator << <>(ostream& outs, const A<T>& object);
	//Other functions of class A
}
#endif //A_H

Implementation file A.cpp

C++

#ifndef A.CPP
#define A.CPP
//the friend function definition and other function definitions
#endif //A.CPP

What I dont understand is, in normal classes you can always define the friend functions after the classes definition. In template classes you either have to define friend function inside the template class or forward declare the friend function. Is there an explanation for this?

What I have tried:

These are the only two ways I know. Can anyone explain how they work and if there is another way?

Posted 21-Aug-16 0:28am

ducdohuunguyen

Updated 21-Aug-16 2:37am

Add a Solution

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Richard MacCutchan · Accepted Answer · 2016-08-21T02:37:00

The reason is because the template is not a class definition, it is a template that is used to create a class definition. It is only when you declare the actual class, which includes the declaration of the type (T) does the compiler actually create the class definition. If the friend declaration is outside the class then there is no guarantee that the type will match what has been declared in the class.