How to declare friend method when class definitions cross reference?

Question

I want to define two classes, A and B. A has a data member which is a Class B object and is in-class initialised. A also has a method to retrieve the value in this B type data member and this method would be declared as a friend method in B. Here is my code:

class A{ public: int getBValue(); private: B b=B(1); }; class B{ public: friend int A::getBValue(); B(int i):value(i){} private: int value; }; int A::getBValue(){ return b.value; }

And unsurprisingly the compilation had failed because of unknown type B in class A definition. I had tried to swap the definitions of A and B in the source and the result was even worse. Is there a possible way to resolve this cross reference issue between A and B?

What was the error when you declared B first? What happens if you only make the class 'A' as friend when switching the order? Anyway, why not have a public getter? — László Papp
– László Papp, Commented Dec 25, 2013 at 12:21
You can't make this without a compromise - either store a pointer to B instead of an instance and forward declare it, or give up on friend requirement and provide a getter instead. — jrok
– jrok, Commented Dec 25, 2013 at 12:27
@LaszloPapp Two errors. One is "use of undeclared identifier 'A'", the other is "'value' is a private member of 'B'". — God_of_Thunder
– God_of_Thunder, Commented Dec 25, 2013 at 13:10

Martijn Courteaux · Accepted Answer · 2013-12-25 12:27:50Z

If this is the complete code as you have it, then the problem is that the compiler doesn't know what a B is at the time it is compiling class A. One way to solve it is by creating a pointer to B instead of having a B itself:

A.h

#ifndef CLASS_A #define CLASS_A class B; class A{ public: int getBValue(); private: B *b; }; #endif

B.h

#ifndef CLASS_B #define CLASS_B #include "A.h" class B{ public: friend int A::getBValue(); B(int i):value(i){} private: int value; }; #endif

A.cpp

#include "A.h" #include "B.h" int A::getBValue(){ return b->value; }

Adrian W · Accepted Answer · 2013-12-25 20:14:49Z

Replacing an embedded member of type B with a pointer (or reference) to a B changes the way your class works and needs additional care when copying objects of class A.

When you reverse the definitions, you cannot make a member function of class A a friend of class B, because the type of A is incomplete at the time of the friend declaration. But you can make the entire class A a friend.

A solution with class A having an embedded member of class B could look like this:

class A; class B{ public: friend class A; B(int i):value(i){} private: int value; }; class A{ public: int getBValue(); private: B b=B(1); }; int A::getBValue(){ return b.value; }

Changes made:

Declared B before A. A forward declaration for A is used.
Made the class A a friend of B. This works even when A is not yet fully defined.

Compiles with g++ version 4.7.2 (-std=c++11 needed for B b=B(1);)

Anyway, having a member of class A accessing a private member of class B is something which can (and should) almost always be avoided (see Laszlo Papp's comment on your post).

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