4

My goal is simple - I want to access the protected members of one class from the methods of another class. For this I have the following -

A.HPP

#ifndef A_HPP
#define A_HPP

#include "B.hpp"
using namespace std;

class A
{
    protected:
        int a;
    public:
        A();
        ~A(){};
        friend void B::geta(A& ao);
};

inline A::A()
{
    a = 2;
    cout << a;
}

#endif

B.HPP

#ifndef B_HPP
#define B_HPP

using namespace std;

class A;

class B
{
    protected:
        int b;
    public:
        B();
        ~B(){};
        void geta(A& ao);
};

inline B::B()
{
    b = 1;
    cout << b;
}

inline void B::geta(A& ao)
{
    b = ao.a;
    cout << b;
}

#endif

MAIN.CPP

#include <iostream>

#include "A.hpp"
#include "B.hpp"

int main()
{
    A ao;
    B bo;
    bo.geta(ao);
    return 0;
}

When I compile this I get the following error.

error

How do I fix this? Most answers that I see here are just putting all classes in one file and defining functions in appropriate positions to achieve this but I need them in separate files.

Papa Delta
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2 Answers2

4

Option 1: Not inline

You can of course move the definition of B::geta to a B.cpp file which includes A.hpp and remove the inline keyword. But this might make compiler optimizations less likely.

Option 2: Weird #include logic

Your definitions as they are can only compile if the compiler sees a forward declaration of A, the definition of B, the definition of A, and the definition of B::geta, all in that order. So if you want the definition of A in one file and the definition of B in another, you'll need to get the preprocessor to do some switching back and forth between files.

A.hpp

// NO A_HPP include guard!
#ifdef INCLUDED_FROM_B_HPP

class A
{
    protected:
        int a;
    public:
        A();
        ~A(){};
        friend void B::geta(A& ao);
};

inline A::A()
{
    a = 2;
    cout << a;
}

#else
#  include "B.hpp"
#endif

B.hpp

#ifndef B_HPP
#define B_HPP

class A;

class B
{
    protected:
        int b;
    public:
        B();
        ~B(){};
        void geta(A& ao);
};

#define INCLUDED_FROM_B_HPP
#include "A.hpp"
#undef INCLUDED_FROM_B_HPP

inline B::B()
{
    b = 1;
    cout << b;
}

inline void B::geta(A& ao)
{
    b = ao.a;
    cout << b;
}

#endif

So now if a file does #include "B.hpp", the preprocessor will:

  1. Output the class A; and definition of B from the first part of B.hpp.
  2. Define INCLUDED_FROM_B_HPP.
  3. Output the definitions in A.hpp.
  4. Clear the definition of INCLUDED_FROM_B_HPP.
  5. Output the definitions of inline members of B from the second part of B.hpp.

If a file does #include "A.hpp" first, things are a bit trickier:

  1. Since INCLUDED_FROM_B_HPP is not set, just immediately go into B.hpp.
  2. Output the class A; and definition of B from the first part of B.hpp.
  3. Define INCLUDED_FROM_B_HPP.
  4. When the #include "A.hpp" in the middle of B.hpp is encountered, the preprocessor will go recursively back into A.hpp. But this time since INCLUDED_FROM_B_HPP is defined, it outputs the code contents of A.hpp.
  5. Clear the definition of INCLUDED_FROM_B_HPP.
  6. Output the definitions of inline members of B from the second part of B.hpp.

Option 3: friend class B;

Instead of specifying just the one member function B::geta to friend, just befriend the class itself, with the declaration friend class B;. Now A.hpp doesn't need to include B.hpp, so there's no circular dependency issue.

This doesn't much decrease encapsulation from the point of view of what access is and isn't possible, since normally any programmer who can modify any part of class B can also modify B::geta. But it does open possibilities of "accidentally" using non-public members of A in other members of B.

Option 4: Refactor access method

A.hpp

#ifndef A_HPP
#define A_HPP

class B;

class A
{
    protected:
        int a;
    public:
        A();
        ~A(){};

        class AccessForB {
        private:
            static int geta(A& aobj) { return aobj.a; }
            friend class ::B;
        };
};

inline A::A()
{
    a = 2;
    cout << a;
}

#endif

B.hpp

...

inline void B::geta(A& ao)
{
    b = A::AccessForB::geta(ao);
    cout << b;
}

This code introduces a new sort of encapsulation: now class B can only get the value from member a, can't modify that value, and can't access any other non-public members of A. Additional accessors could be added for other members as appropriate. To permit modifying a member, the class could either provide "set" accessors, or an accessor that returns a reference. For non-public functions, the class could provide wrapper functions that just pass through to the actual function.

It's still true that it's possible for members of B other than B::geta to exploit the friendship, but now typing out A::AccessForB:: can't really be considered an accident.

aschepler
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0

... but I need them in separate files.

Move the inlined functions out to a separate B.cpp file that #include's A.hpp.

As being just seen as a forwarded class declaration A's declaration isn't yet complete at the point you're starting to use it in B.hpp, and the compiler complains about that righteously.

You can still retain the inline keyword, the compiler will respect it as a hint as usual.

Here's the sketch:

B.hpp

#ifndef B_HPP
#define B_HPP

using namespace std;

class A;

class B
{
    protected:
        int b;
    public:
        B();
        ~B(){};
        void geta(A& ao);
};
#endif

B.cpp

#include "B.hpp"
#include "A.hpp"

inline B::B()
{
    b = 1;
    cout << b;
}

inline void B::geta(A& ao)
{
    b = ao.a; // <<<<<<<<<<<< here's the problem to be solved
    cout << b;
}

These questions and answers are very much related to your problem:

πάντα ῥεῖ
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  • Can I not move the definitions to a `hpp` file say `A_impl.hpp` and `B_impl.hpp` instead? I will have lot of these files and I don't want to generate avoidable object files? – Papa Delta Sep 14 '18 at 04:36
  • @User525412790 No you can't. All `#include` statements will be resolved prior to any actual compilation by the preprocessor. The linking stage might respect `ìnline` keywords as much as possible thoogh. – πάντα ῥεῖ Sep 14 '18 at 04:38
  • This won't work, assuming some other translation unit will need to odr-use one of those member functions. – aschepler Sep 14 '18 at 04:58
  • @aschepler What do you mean _won't work_? inlining the functions in question? – πάντα ῥεῖ Sep 14 '18 at 05:01
  • @ πάντα ῥεῖ : Were you able to execute it? My A and B cpp files compile and then I create a `.a` file with those but I get linking errors when I try to execute `main()` – Papa Delta Sep 14 '18 at 05:01
  • An inline function must be defined in every translation unit in which it is odr-used. – aschepler Sep 14 '18 at 05:03
  • @User525412790 Did you manage to link `B.o` with your final output properly? – πάντα ῥεῖ Sep 14 '18 at 05:03
  • @aschepler OK, that's well answered in the additional links I mentioned. If it can't be inlined for these reasons, it can't be inlined. Isn't that a minor point? – πάντα ῥεῖ Sep 14 '18 at 05:06
  • @πάνταῥεῖ I'm not talking about "can be inlined". I'm talking about an ill-formed program, which explains the link-time error. – aschepler Sep 14 '18 at 05:07