如何从一个基类中公开继承，但在派生类中使一些来自基类的公共方法私有？ [英] How to publicly inherit from a base class but make some of public methods from the base class private in the derived class?

问题描述

例如， Base 有两个公共方法： foo（）和 bar （）。类派生从类 Base 继承。在 Derived 中，我想让 foo（） public但 bar code> private。以下代码是否正确和自然地执行此操作？

For example, class Base has two public methods: foo() and bar(). Class Derived is inherited from class Base. In class Derived, I want to make foo() public but bar() private. Is the following code the correct and natural way to do this?

class Base {
   public:
     void foo();
     void bar();
};

class Derived : public Base {
   private:
     void bar();
};

推荐答案

C ++ '03标准的第11.3节描述了这种能力：

Section 11.3 of the C++ '03 standard describes this ability:

11.3访问声明

基类的
成员的访问权限可以更改
在派生类中声明它的
限定id在派生类
声明。这种提及被称为
访问声明。
访问声明的效果qualified-id;是
，定义为等同于使用qualified-id的
声明

11.3 Access declarations
The access of a member of a base class can be changed in the derived class by mentioning its qualified-id in the derived class declaration. Such mention is called an access declaration. The effect of an access declaration qualified-id ; is defined to be equivalent to the declaration using qualified-id

注意：从ISO C ++ '11开始，访问声明（ Base :: bar; / code>）被禁止，如评论中所述。应该使用一个使用声明（使用Base :: bar; ）。

1）您可以使用公共继承，然后将栏设为私人：

1) You can use public inheritance and then make bar private:

class Base {
public:
    void foo(){}
    void bar(){}
};

class Derived : public Base {
private:
    using Base::bar;
};

2）可以使用私有继承，然后使foo public：

2) You can use private inheritance and then make foo public:

class Base {
public:
    void foo(){}
    void bar(){}
};

class Derived : private Base {
public:
    using Base::foo;
};

注意：如果你有一个类型为Base的指针或引用，其中包含Derived类型的对象，用户仍然可以调用该成员。

Note: If you have a pointer or reference of type Base which contains an object of type Derived then the user will still be able to call the member.

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