默认类继承访问 | 默认类继承访问

本文介绍了默认类继承访问的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

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：
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:
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.
                        这篇关于默认类继承访问的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！