为什么我的虚拟方法没有被覆盖? [英] Why my virtual method is not overridden?

问题描述

class Base
{
public:
Base()
{
cout<<"base class"<<endl;
fun();
}
virtual void fun(){cout<<"fun of base"<<endl;}
};

class Derive:public Base
{
public:
Derive()
{
cout<<"derive class"<<endl;
fun();
}
void fun(){ cout<<"fun of derive"<<endl;}
};

void main()
{
Derive d;
}

输出为:

base class
fun of base
derive class
fun of derive

为什么第二行不是fun of derive?

推荐答案

在基类构造函数中调用fun()时，尚未构造派生类(在C ++中，类首先是构造的父类)，因此系统尚无Derived实例，因此在Derived::fun()的虚拟功能表中没有条目.

When you call fun() in the base class constructor, the derived class has not yet been constructed (in C++, classes a constructed parent first) so the system doesn't have an instance of Derived yet and consequently no entry in the virtual function table for Derived::fun().

这就是为什么通常不赞成在构造函数中调用虚拟函数的原因，除非您明确地要调用虚拟函数的实现，该实现是当前实例化对象的一部分或一部分它的祖先.

This is the reason why calls to virtual functions in constructors are generally frowned upon unless you specifically want to call the implementation of the virtual function that's either part of the object currently being instantiated or part of one of its ancestors.

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