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？

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();
};

答

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.

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？

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