friend class doesn't work when I put in namespace?

Winter Knight · 12-12-2007, 02:23 AM

Here is a simple c++ program that demonstrates friend classes:

Code:

class A {
friend class B;

private:
	int var;
};

class B {
public:
	int GetVar() { return myA.var; }

private:
	A myA;
};

int main() {}

The above code compiles just fine. However, when I put class A into a namespace, like so:

Code:

namespace wk {

class A {
friend class B;

private:
	int var;
};

} // namespace wk

class B {
public:
	int GetVar() { return myA.var; }

private:
	wk::A myA;
};

int main() {}

The above code does not compile. I get the following error message:

Code:

wk@ALPHA:~/cprog$ g++ -o tmp tmp.cpp
tmp.cpp: In member function 'int B::GetVar()':
tmp.cpp:7: error: 'int wk::A::var' is private
tmp.cpp:14: error: within this context

I can put class A in a namespace, but only if I make var public, like so:

Code:

namespace wk {

class A {
public:
	int var;
};

} // namespace wk

class B {
public:
	int GetVar() { return myA.var; }

private:
	wk::A myA;
};

int main() {}

The above code compiles just fine, without error.

Does anyone know why this is? It looks like using namespaces nullifies friend classes. I couldn't find any mention of this in any of the c++ references that I found on the internet. Is there a way to keep A.var private, while keeping it in a namespace?

Looking_Lost · 12-12-2007, 03:00 AM

Only learning C++ myself, but I found if you stick a forward declaration of class B at the top of your file it will compile, scope issue maybe ?

Code:

class B;

namespace wk {

class A {
friend class B;

private:
	int var;
};

} // namespace wk

class B {
public:
	int GetVar() { return myA.var; }

private:
	wk::A myA;
};

int main() {}

Winter Knight · 12-12-2007, 07:16 AM

The code you displayed does not compile for me. I get the same error. What compiler are you using? I'm using Debian stock kernel, g++ 4.2.2-4.

Looking_Lost · 12-12-2007, 07:25 AM

On a windows machine at the moment using CYGWIN environment, gnu g++ 3.4.4

Winter Knight · 12-12-2007, 07:41 AM

I just downloaded, installed, and tried g++-3.4 on your version of the program, and I get the same results as you. I wonder which one is correct. I still don't understand why my original doesn't compile. Their shouldn't be a scope issue, because the code inside a block, such as the "wk" namespace, should be able to see outside of it's block. It's just that nobody should be able to see inside without fully qualifying (which I do).

This might be a compiler bug, but I'm not sure. I've thought that before, and then someone comes along and shows me my error.

ntubski · 12-12-2007, 08:19 AM

The following works for me on g++ version 4.2.3

Code:

class B;

namespace wk {

class A {
friend class ::B;

private:
	int var;
};

} // namespace wk

class B {
public:
	int GetVar() { return myA.var; }

private:
	wk::A myA;
};

int main() {}

Winter Knight · 12-12-2007, 08:25 AM

Thanks. That works for me too. It still seems odd, though.

dmail · 12-12-2007, 10:39 AM

ntubski reply is correct and it is due to scope(namespace issue), by leaving out the scope operator you are saying the class is in the same namespace as the other class. ie

Code:

namespace BAR{ class Bar{};}
class Bar{};
namespace FOO{ 
class Bar{};
class Foo
{
friend class Bar;//says Bar is really FOO::Bar
firend class ::Bar;//says Bar is a member of the global namespace
firend class BAR::Bar;//says Bar is a member of the BAR namespace
}
; 
}

plus you have to forward declare the class.

Winter Knight · 12-12-2007, 08:20 PM

OK. Thanks. I get the part about scope. But why do you have to forward declare only if you use a namespace, but not if the two classes are in the same namespace?

For example, this works:

Code:

class A {
friend class B;

private:
	int var;
};

class B {
public:
	int GetVar() { return myA.var; }

private:
	A myA;
};

int main() {}

But this doesn't:

Code:

namespace wk {

class A {
friend class ::B;

private:
	int var;
};

} // namespace wk

class B {
public:
	int GetVar() { return myA.var; }

private:
	wk::A myA;
};

int main() {}

In the second example, of course, it works if I forward declare class B.

paulsm4 · 12-12-2007, 11:35 PM

Strictly speaking, this is the correct behavior. Here's why:

http://womble.decadentplace.org.uk/c...ax-errors.html

Here's a more detailed explanation:

http://www.open-std.org/jtc1/sc22/wg...000/n1229.html

And here's a very short explanation:

Quote:

You have to declare friendship to the function in the namespace, not
to a non-existent global one.
...
Just to be accurate, you'll get a bunch of errors since
1) mynamespace is undeclared at this point
2) the Bar declaration has no return value

http://www.thescripts.com/forum/thread62918.html

This is a good example (one of many) why I think C++ is a *terrible* language for beginners - there are just too many subtle "surprises" for anybody but advanced C++ programmers to have to deal with.

IMHO ... PSM

Winter Knight · 12-13-2007, 12:57 PM

Thanks for the links, paulsm4. It helped me understand the situation a lot. Here is an especially interesting quote from the second link:

"The wording in the standard regarding the handling of names declared in unqualified friend declarations is unclear and, in certain cases, clearly incorrect."