Pimpl成语与继承 [英] Pimpl idiom with inheritance

问题描述

我想使用带继承的pimpl习语。

I want to use pimpl idiom with inheritance.

这是基础公共类及其实现类：

Here is the base public class and its implementation class:

class A
{
    public:
      A(){pAImpl = new AImpl;};
      void foo(){pAImpl->foo();};
    private:
      AImpl* pAImpl;  
};
class AImpl
{
    public:
      void foo(){/*do something*/};
};

我希望能够使用其实现类创建派生公共类：

And I want to be able to create the derived public class with its implementation class:

class B : public A
{
    public:
      void bar(){pAImpl->bar();};    // Can't do! pAimpl is A's private.
};        

class BImpl : public AImpl
{
    public:
      void bar(){/*do something else*/};
};

但是我不能在B中使用pAimpl，因为它是A的私有。

But I can't use pAimpl in B because it is A's private.

所以我看到了一些方法来解决它：

So I see some ways to solve it:

1. 在B中创建BImpl * pBImpl成员，并通过它带有附加的A构造函数A（AImpl *）。


2. 将pAImpl更改为受保护（或添加Get函数），并在B中使用它。


3. B不应该继承自A.在B中创建BImpl * pBImpl成员，并在B中创建foo（）和bar（），这将使用pBImpl。


4. 任意其他方式？

1. Create BImpl* pBImpl member in B, and pass it to A with additional A constructor, A(AImpl*).
2. Change pAImpl to be protected (or add a Get function), and use it in B.
3. B shouldn't inherit from A. Create BImpl* pBImpl member in B, and create foo() and bar() in B, that will use pBImpl.
4. Any other way?

我应该选择什么？

推荐答案

class A
{
    public:
      A(bool DoNew = true){
        if(DoNew)
          pAImpl = new AImpl;
      };
      void foo(){pAImpl->foo();};
    protected:
      void SetpAImpl(AImpl* pImpl) {pAImpl = pImpl;};
    private:
      AImpl* pAImpl;  
};
class AImpl
{
    public:
      void foo(){/*do something*/};
};

class B : public A
{
    public:
      B() : A(false){
          pBImpl = new BImpl;
          SetpAImpl(pBImpl);
      };
      void bar(){pBImpl->bar();};    
    private:
      BImpl* pBImpl;  
};        

class BImpl : public AImpl
{
    public:
      void bar(){/*do something else*/};
};

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