Ambiguity in static polymorphism [duplicate]

问题

This question already has answers here:

name hiding and fragile base problem (3 answers)

Closed 4 years ago.

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

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();
}

回答1:

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.

来源：https://stackoverflow.com/questions/32655484/ambiguity-in-static-polymorphism