C++ – How to forward declare a member function of a class to use in another class

cforward-declaration

I have made two identical classes X and Y, with a pointer to each other. See the code below for X.h, Y.h is identical with all X's and Y's interchanged. This code gives however an error in my method Connect (error C2027: use of undefined type 'Y'). In X.h, I have forward declared the class Y, but it doesn't know that Y has a method named SetXPointer. Therefore I also need to forward declare this method, correct?

If I try to do this (adding the line Y::SetXPointer(X* pX_in); under the line class Y;), I get a compiler error C2761: 'void Y::SetXPointer(X *)' : member function redeclaration not allowed. Is there a way to use a public method of class Y in class X?

// X.h

#pragma once

#include "Y.h"

// Forward declaration
class Y;

class X
{
public:
    X(void) : data(24) {};
    ~X(void) {};
    int GetData() { return data; }
    void SetYPointer(Y* pY_in) { pY = pY_in; }
    Y* GetYPointer() { return pY; }
    void Connect(Y* Y_in) { pY = Y_in; Y_in->SetXPointer(this); }
private:
    int data;
    Y *pY;
};

Best Answer

Don't include the method body in the class body. Write both classes, and after both classes are complete, write the method implementations:

class Y;
class X {
  …
  void Connect(Y* Y_in);
  …
};
class Y {
  …
  void Connect(X* X_in);
  …
};
inline void X::Connect(Y* Y_in) {
  pY = Y_in;
  Y_in->SetXPointer(this);
}
inline void Y::Connect(X* X_in) {
  pX = X_in;
  X_in->SetXPointer(this);
}

That way, full information about how the objects of the class will be layed out in memory is available by the time the Connect method is implemented. And as a method in the class body and a method declared inline will both be inlined the same way, performance will be the same as well.

The only downside is that you won't be able to split these two classes over two headers in a reasonable way.

Related Solutions

C++ – When should you use a class vs a struct in C++

The differences between a class and a struct in C++ is:

struct members and base classes/structs are public by default.
class members and base classes/struts are private by default.

Both classes and structs can have a mixture of public, protected and private members, can use inheritance and can have member functions.

I would recommend you:

use struct for plain-old-data structures without any class-like features;
use class when you make use of features such as private or protected members, non-default constructors and operators, etc.

C++ – Templated check for the existence of a class member function

Yes, with SFINAE you can check if a given class does provide a certain method. Here's the working code:

#include <iostream>

struct Hello
{
    int helloworld() { return 0; }
};

struct Generic {};    

// SFINAE test
template <typename T>
class has_helloworld
{
    typedef char one;
    struct two { char x[2]; };

    template <typename C> static one test( decltype(&C::helloworld) ) ;
    template <typename C> static two test(...);    

public:
    enum { value = sizeof(test<T>(0)) == sizeof(char) };
};
    
int main(int argc, char *argv[])
{
    std::cout << has_helloworld<Hello>::value << std::endl;
    std::cout << has_helloworld<Generic>::value << std::endl;
    return 0;
}

I've just tested it with Linux and gcc 4.1/4.3. I don't know if it's portable to other platforms running different compilers.

Best Answer

Related Solutions

C++ – When should you use a class vs a struct in C++

C++ – Templated check for the existence of a class member function

Related Topic