C ++自动实例化派生类 [英] C++ Automatic instantiation of derived classes

问题描述

我有一个名为 Base 的抽象基类，其他程序员将为其编写实现。在应用程序的其他部分，我想捕获所有已编写的实现，并为每个实现构造一个实例。如果可以在实现基础之外没有其他指令的情况下完成此操作，那就太好了。但是，我下面的代码要求每个实现自己注册。

I have an abstract base class called Base that other programmers are to write implementations for. In some other part of the application, I want to catch all implementations that have been written and construct a single instance of each. If this could be done with no additional instructions to others beyond "implement Base", that would be beautiful. However, the code I have below, requires that each implementation register itself. It also doesn't work.

#include <iostream>
#include <vector>

class Base;

std::vector<Base*>* registrationList = new std::vector<Base*>;

class Base {
public:
   Base(){}
   virtual void execute() = 0;
};

class ImplementationOne: public Base {
public:
   ImplementationOne(){registrationList->push_back(this);}
   void execute(){std::cout << "Implementation One." << std::endl;}
   static int ID;
};

class ImplementationTwo: public Base {
public:
   ImplementationTwo(){registrationList->push_back(this);}
   void execute(){std::cout << "Implementation Two." << std::endl;}
   static int ID;
};

int main(int argc, const char * argv[]){
   std::cout << "Registration List size: " << registrationList->size() << std::endl;
   for(auto it = registrationList->begin() ; it != registrationList->end() ; ++it){
      (dynamic_cast<Base*>(*it))->execute();
   }
   return 0;
}

我得到以下输出：注册列表大小： 0 ，因此很明显这些实现从未实例化。很可能不会发生这种情况，但是我是一个初学者，这是我能想到的最好的方法。我假设 static int ID; 将强制实例化每个实现，然后将其注册。我可以看到 static 不会导致实例化。我将其留在代码中，因为它显示了我的意图。

I get an output of: Registration List size: 0, so it is clear that the implementations were never instantiated. It is probably obvious that this wouldn't happen, but I am a beginner and this is the best I could come up with. I assumed that static int ID; would force instantiation of each implementation, which would then register themselves. I can see static does not result in instantiation. I leave it in my code here since it shows my intent.

我该怎么做才能自动实例化每个实现？

What can I do to get automatic instantiation of each implementation? Is it possible?

推荐答案

添加 static 成员不会 导致实例的生成，它仅声明此类型具有全局变量。但是，您实际上从未定义过这些成员，因此，如果您尝试将它们用于任何用途，则将出现链接器错误。您必须实际实例化一个对象并将其注册。

Adding static members does not cause an instance to be generated, it merely declares that this type has a "global" variable. You never actually defined these members though, so if you tried to use them for anything you would have had a linker error. You'll have to actually instantiate an object and register that.

一种解决方案可能是简单地要求每种派生类型在启动时注册实例。正如我在此处所示，这实际上很容易做到。 （请注意，我已将全局变量移到静态函数的静态局部。这可以防止您尚未遇到的一些问题，包括为全局变量提供所有者。）

One solution might be to simply require each derived type to register an instance at startup. This is actually quite easily done, as I show here. (Note I moved your global to a static local of a static function. This prevents several problems you haven't run into yet, including providing an "owner" for the global.)

---

与您的问题无关，您的代码有问题：

---

Unrelated to your issue, your code has problems:

• 几乎没有任何理由


• 您是从没有虚拟析构函数的类型中多态派生的。


• 您 dynamic_cast< Base *> ，没有任何正当理由。


• 每个派生类都声明但未定义 ID 成员。

• There is almost never any reason to have a pointer to a container.
• You are deriving polymorphically from a type with no virtual destructor.
• You dynamic_cast<Base*> for no apperent reason.
• Each of your derived classes declares but does not define a ID member.

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