以编程方式获取派生类的名称 [英] Programmatically getting the name of a derived class

问题描述

我正在尝试做类似的事情:

I am attempting to do something like:

class Base {
public:
   Base() {
      cout << typeid(*this).name() << endl;
   }
   ...
};

class Derived : public Base { ... }
class MoreDerived : public Derived { ... }

Derived d;
MoreDerived m;

问题是，当我需要查看Derived和MoreDerived时，总是将Base打印到屏幕上.有没有一种方法可以让Typeid与派生类一起使用?还是除了typeid之外还有另一种方法?

Problem is, I always get Base printed to the screen, when I need to see Derived and MoreDerived. Is there a way to get typeid to work this way with derived classes? Or is there another approach besides typeid?

注意:我要向已经编码的套件中添加功能，所以我不想在基类中添加虚拟方法，派生类自己返回该值.另外，不必担心运行时的开销，这将成为调试编译开关的一部分.

Note: I am adding functionality to an already coded suite, so I don't want to have to add a virtual method to the base class where the derived classes return this value themselves. Also, not worried about runtime overhead, this will be part of a debug compile switch.

推荐答案

在构造函数Base()中，该对象仍然是"Base"实例.在Base()构造函数之后，它将成为Derived实例.尝试在构造完成后 ，它将起作用.

In the constructor Base(), the object is still a "Base" instance. It will become a Derived instance after the Base() constructor. Try to do it after the construction and it will work.

例如:

避免在构造函数中使用虚拟方法

http://www.artima.com/cppsource/nevercall.html

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