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Non-local class, namespace scope, function name must be unqualified.

Hello Experts!

Found such a description and I do not really understand it...

However, you can define a function in a friend declaration. The class must be a non-local class. The function must have namespace scope, and the function name must be unqualified. The following example demonstrates this:

1. what is non-local class?
2. The function must have namespace scope- what does it mean?
3. Function name must be unqualified- what does it mean?


class A {
  void g();
};

void z() {
  class B {
    friend void f() { }; // error
  };
}

class C {
  friend void A::g() { } // error
  friend void h() { }
};

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Thank you!
0
panJames
Asked:
panJames
  • 3
1 Solution
 
ambienceCommented:
1. what is non-local class?

A class declared inside a function or method is non-local. It can not be used outside the defining scope.
0
 
ambienceCommented:
3. Function name must be unqualified?

These are qualified function names

  friend void ::h() {
  }
  friend void Test::h() {
  }

This is not

  friend void h() {
  }

WRT - friends that this means is

void f() {}

namespace Test {

      void f() {}

      class C
      {
                // definition must be unqualified
            friend void h() {
                  cout << "From h" << std::endl;
            }
            friend void Test::f(); // qualified declaration is allowed
            friend void ::f(); // qualified declaration is allowed

      public:
            void method()
            {
                  h();
            }
      };

}
0
 
ambienceCommented:
2. The function must have namespace scope- what does it mean?

This should explain

namespace Test {

	void h();  // h is declared in namespace - not defined!

	class C 
	{
	public:
                // this is inside C but its definition of h in the namespace's scope
		friend void h() { 
			cout << "From h" << std::endl;
		}
	};

}


int main() 
{
	Test::h();
}

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