friend problem

[lukak]

Hi, I think I have misunderstood friend keyword. I'm trying following:

class X
{
friend class Y;
friend void f(){}
};

class Z
{
Y *p;
void g(){ return ::f();}
};

int main()
{ return 0; }

error: ISO C++ forbids declaration of ‘Y’ with no type
‘::f’ has not been declared

Friend keyword puts class Y to outer scope so class Y should be global declared class and available in class Z. Another problem is with friend function which is defined in class Y, so function call f() in Z class should succeed. Reference: 466 page C++ primer 4th edition (Friend declaration and scope).

Anyone has a explanation why program wont compile?
Sorry if my English is not good as it should be...

[jsmith]

Well, nowhere have you shown a declaration for class Y. You've shown declarations for X and Z.

[lukak]

To my understanding, only friend member functions require class definition to be present before friend declaration. Friend class-es and non-member functions need not to be declared to be made a friend. It is written that the scope of friend declaration inside function is exported to enclosing scope. So class Y and function f() are in the global scope in this case, so it should be possible to use them in class Z (pointer and references of declared non-defined classes persist). Alias:

class Y;
void f(){}

class X
{
.
.
.
Words of wisdom appreciated...

[jsmith]

The first problem is that in order to declare a variable of type Y* (p member in Z), you must at least forward declare Y.

Second, saying

friend void f(){}

declares void f() to be a friend of X, but it does not actually declare void f(). This compiles for me:

class X
{
friend class Y;
friend void f();     // Removed implemention from here;  not a declaration of f().
};

void f(){}        // Added declaration of f() and implementation of f()

class Y;
class Z
{
Y *p;
void g(){ return ::f();}
};

int main()
{ return 0; }

[lukak]

Yes I understand that but I don't understand friend keyword.....I quote : "The scope of the function is exported to the scope enclosing the class definition". The example I have given is copied from the book with comment that Y class in Z class is introduced by friend in x. No forward declaration is given. Further more the definition of the function f as friend in class X isn't declared globally. So I think the best solution would be to explain in detail friend keyword. I would appreciated that....

[jsmith]

You understand the keyword friend better than the book. The book you have is wrong. friend void f(); simply says that void f() is a friend of X -- nothing more. It is simply a statement of friendship. It does not serve as a declaration of the function f(). In my code above, line 13 references f(). It is line 7, not line 4, that declares f().

[lukak]

Thank you for your trouble, reps where helpful....I will try to find another explanation before I question the book. But it seems that keyword isn't that kind of mystery after all. TNX...