默认类继承访问 [英] Default class inheritance access

问题描述

假设我有一个基础和派生类：

  class Base 
 {
 public：
 virtual void Do（）; 
} 
 
 class Derived：Base 
 {
 public：
 virtual void Do（）; 
} 
 
 int main（）
 {
 Derived sth; 
 sth.Do（）; // calls Derived :: Do OK 
 sth.Base :: Do（）; // ERROR;不调用Based :: Do 
} 
  

如果我希望访问Base :: Do通过Derived。当我声明Derive为

 时，我得到一个编译错误class Base in unrecessibleclass Derived：public Base $ b $ 
 
 
 
我已读取默认继承
解决方案
您可能读过不完整或误导的内容。引用Bjarne Stroustrup从C ++编程语言，第四版，p。 602：
 
 
在类中，成员默认为 private ;在 struct 中，成员
默认为 public （§16.2.4）。
 
 
这也适用于没有访问级别说明符的继承成员。
 
 
 
只使用struct来组织纯数据成员。您正确地使用了类来建模和实现对象行为。
 
Suppose I have a base and derived class:
class Base
{
    public:
    virtual void Do();
}

class Derived:Base
{
    public:
    virtual void Do();
}

int main()
{
    Derived sth;
    sth.Do(); // calls Derived::Do OK
    sth.Base::Do(); // ERROR; not calls Based::Do 
}
as seen I wish to access Base::Do through Derived. I get a compile error as "class Base in inaccessible" however when I declare Derive as 
class Derived: public Base
it works ok.

I have read default inheritance access is public, then why I need to explicitly declare public inheritance here?
解决方案
You might have read something incomplete or misleading. To quote Bjarne Stroustrup from "The C++ programming Language", fourth Ed., p. 602:

  
In a class, members are by default private; in a struct, members
  are by default public (§16.2.4).
This also holds for members inherited without access level specifier.

A widespread convention is, to use struct only for organization of pure data members. You correctly used a class to model and implement object behaviour.

                        
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