C ++同名的函数的继承 [英] C++ Inheritance of functions with same name

问题描述

我有以下类声明：

  class human 
 {
 public：
 void msg（）{cout<<I am human\\\
;} 
}; 
 class John：public human 
 {
 public：
 void msg（）{cout<<我是John \\\
;} 
}; 
  

显然， John human 都有一个函数 msg（）。显然， John 类继承了来自 human 的资源。现在，当我创建一个派生类的对象并调用 msg（）：

  John a; 
 a.msg（）; 
  

输出为：

我是John

但不会 John 继承 msg（）从人？

也有一种方法来访问人 msg（） $ c>通过使用派生类的对象？

编辑：

这样会帮助我

  a.human :: msg（）
 另一个问题：
 
 
 
另外一个问题：如果我修改类，像这样：pre 
 
 < 
  class human 
 {
 protected：
 void msg（）{cout< I am human\\\
;} 
}; 
 class John：public human 
 {
 public：
 void msg（）{cout<<我是John \\\
;} 
}; 
  
现在我如何访问 msg >  
解决方案
 
 em>但是约翰不从人类继承msg（） 
 
 
自己的实现。
 
 
 还有一种方法来访问人类的msg > 
 
是的，需要显式声明类范围（长 human :: msg（）在公共范围内，因为它最初被问到）：
  John a; 
 a.human :: msg（）; 
 // ^^^^^^^ 
  
需要应用相同的语法if您希望从类中访问 human :: msg（）：John ：
  class John：public human {
 public：
 void msg（）{cout< } 
 void org_msg（）{human :: msg（）; } 
}; 
 
 John a; 
 a.org_msg（）; 
  
或另一个替代
  John john; 
 human * h =& john; 
 h-> msg（）; 
  
 
I have the following class declarations:
class human
{
    public:
    void msg(){cout<<"I am human\n";}
};
class John:public human 
{
    public:
    void msg(){cout<<"I am a John\n";}
};
As it is clear the class John and human both have a function msg(). Clearly , the class John is inheriting resources from human. Now when I create an object of the derived class and call msg():
John a;
a.msg();
The output is :

  
I am John
But doesnt John inherit msg() from human? 


Also is there a way to access msg() of human by using an object of the derived class? 

EDIT:

Yup calling the function like so will help me
a.human::msg()
Another question :

Also if I modify the class like so:
class human
{
    protected:
    void msg(){cout<<"I am human\n";}
};
class John:public human 
{
    public:
    void msg(){cout<<"I am a John\n";}
};
Now how can I access the msg() of human.
解决方案

  
"But doesnt John inherit msg() from human?"
Yes it does, but hides it with its own implementation.

  
"Also is there a way to access msg() of human?"
Yes, you need to explicitly state the class scope (as long human::msg() is in public scope as it was originally asked):
  John a;
  a.human::msg();
 // ^^^^^^^
The same syntax needs to be applied if you want to access human::msg() from within class John:
class John : public human {
    public:
    void msg() {cout<<"I am a John\n";}
    void org_msg() { human::msg(); }
};

John a;
a.org_msg();
or another alternative
John john;
human* h = &john;
h->msg();


                        
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