根据条件将基类类指针指向几个可能的派生类型指针之一 [英] cast base class pointer to one of several possible derived type pointer based on condition

问题描述

我有一个基类 B 和几个派生模板类 D< int> ， D< float> ， D< double> 等等（超过十个）

在我的程序中，我发现一个情况，我有一个B指针，我知道指向一个实例的D专业化。我也有一个唯一的键来标识派生类型。

所以，我想使用我的基类指针和唯一的类型键来调用正确的派生类方法。我实际上想在几个地方做到这一点，所以我提出的一个解决方案不仅是丑陋，而且麻烦。

有更好的方法来访问在C ++中有基指针和唯一键的派生类成员吗？
$ b

我不想/不能修改基类。 boost库是公平的游戏。

这是我的潜在代码，但我不喜欢复制这个到处需要访问一个派生成员函数/变量。所有这一切用于一个成员函数调用？

  B * bptr = ... 
 std :: type_info * typekey = ... 
 
 if（typekey ==& typeid（D< float>））{
 D< float> * dptr = static_cast< D< float> *>（bptr）; 
 dptr-> derivedMember（）; 
} else if（typekey ==& typeid（D< double>））{
 D< float> * dptr = static_cast< D double> *>（bptr）; 
 dptr-> derivedMember（）; 
} else if（typekey ==& typeid（D< int>））{
 D< float> * dptr = static_cast< D< int> *>（bptr）; 
 dptr-> derivedMember（）; 
} 
  

解决方案

$ c> D< type> :: 方法具有相同的名称'DerivedMember'，我假设基类B不声明这是虚拟的，你可以更改你的hierachy：

  class B {etc.}; 
 
 template< class T> 
 D类：public B 
 {
}; 
  

到：

 code> class B {etc.}; 
 
 class C：public B 
 {
 virtual void derivedMember（）= 0; 
}; 
 
 
 template< class T> 
 class D：public C 
 {
 public：
 void derivedMember（）{etc.}; 
}; 
  

然后您可以：

  void f（C * c）/ *或取一个B *和do：C * c = static_cast  * / 
 {
 c-> derivedMember（）; 
} 
  

I have a base class B and several derived template classes D<int>, D<float>, D<double>, etc. (so more than ten)

In my program, I find a situation where I have a B pointer that I KNOW points to an instance one of the D specializations. I also have a unique key that identifies the derived type.

So, I want to call the correct derived class method using my base class pointer and the unique type key. I actually want to do this in several places, so the one solution I have come up with is not only ugly, but cumbersome.

Is there a better way to access the derived class's members having a base pointer and a unique key in C++?

I don't want to / can't modify the base class. boost library is fair game.

Here is my potential code, but I'm not happy with replicating this everywhere I need to access a derived member function/variable. All of this for one member function call?!!

B * bptr = ...
std::type_info * typekey = ...

if        (typekey == &typeid(D<float>) ) {
    D<float> * dptr = static_cast<D<float>*>(bptr);
    dptr->derivedMember();
} else if (typekey == &typeid(D<double>) ) {
    D<float> * dptr = static_cast<D<double>*>(bptr);
    dptr->derivedMember();
} else if (typekey == &typeid(D<int>) ) {
    D<float> * dptr = static_cast<D<int>*>(bptr);
    dptr->derivedMember();
} 

解决方案

If all the D<type>:: methods have the same name 'DerivedMember`, and I assume that the base class B does not declare this as virtual, can you change your hierachy from:

class B { etc. };

template<class T>
class D : public B
{
};

to:

class B { etc. };

class C : public B
{
 virtual void derivedMember() = 0;
};


template<class T>
class D : public C
{
public:
void derivedMember() { etc. };
};

Then you could have:

void f(C* c) /* or take a B* and do: C* c = static_cast<C*>(b); */
{    
  c->derivedMember();
}

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