静态多态性的模糊性 [英] Ambiguity in static polymorphism

问题描述

这是一个简单的问题，但实际上是一个bug，我很多，我无法找到一个合理的答案。

Quite a simple question actually but one which is bugging me a lot and I'm unable to find a sound answer for it.

对于以下代码，函数调用 d.show（）; 在 b =& d; 之后的 b-> show（）; Base _ :: show（）？

How is the function call d.show(); NOT ambiguous for the following code AND why does b->show(); after b = &d; leads to the calling of Base_::show()?

#include <iostream>

using std::cout;

class Base_
{

     public:
          void show()
          {
               cout<<"\nBase Show";
          }
};

class Derived_ : public Base_
{
     public:
          void show()
          {    
              cout<<"Derived Show"<<endl;
          }     
};

int main()
{
     Base_ b;
     Derived_ d;
     b->show(); 
     d.show();
}

推荐答案

在C ++中。基本上，如果一个类定义了一个函数，它隐藏所有来自父类的具有相同名称的函数。还值得注意的是，如果你有一个 Base _ * 一个派生对象，并调用该函数，它会调用基类版本，因为它不是虚拟的，不会尝试在派生类中找到覆盖的实现。

This is called "Name Hiding" in C++. Essentially, if a class defines a function, it hides all functions with the same name from parent classes. Also worth noting, if you had a Base_* to a derived object, and called that function, it would call the base classes version as it is not virtual and will not attempt to find overrided implementations in derived classes.

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