Question

项目A， PA.h ：

#include "Common.h"

class PA
{
   void func()
   {
     Common::getInstance()->cm();
     Common::getInstance()->onlyCallByPA();
   }
}

Project Common， Common.h ：

class Common
{
    SINGLETON
public:
  void cm(){}
 private:
    //I do not want PB to call onlyCallByPA
    //so I want to add class PA to friend class
    //so I need to include PA.h
    //but if I include PA.h, PB.cpp  include PA.h 
    //this will make PA.h expose to PB
    //I do not want PB to include PA.h
  void onlyCallByPA(){}
}

项目B， PB.cpp ：

#include "Common.h"

class PB
{
    //I need to call cm() but PB do not be allowed to call onlyCallByPA
    //and also do not be allowed to include PA.h
}

所以我想将PA作为Common的朋友班，但这会引入对PB的依赖。
更好的解决方案？或者，我可以使用其他设计来实现我想要的吗？

Answer 1

使用前瞻性声明。这将允许您声明友谊，而不包括标题或依赖PA的完整类声明。

COMMON.H

class PA; // forward declaration.

class Common
{
    friend PA;
};

Answer 2

你的意思是：

friend return_type class_name::function(args);

class Common 
{

    // ...

    friend void PA::OnlyCallByPA(); // Only PA has access
};

// EDIT

为PA创建基类，并仅在“Common.h”中引入基类

如何使用朋友类但引入较少的依赖性

2 个答案: