您可以坚持最严格和最纯粹的OOP原则，并确保任何类的数据成员都没有访问器，这样所有对象都必须是唯一可以知道它们的数据的对象，并且对它们进行操作的唯一方法是通过间接消息(即方法)。

但即使是c#也有一个内部可见性关键字，Java也有默认的包级可访问性。c++实际上更接近于OOP的理想，它通过精确地指定哪些其他类或只有其他类可以看到一个类，从而最大限度地减少了类的可见性。

我不太使用c++，但如果c#有朋友，我会用它来代替我经常使用的汇编全局内部修饰符。它并没有真正打破封装，因为。net中的部署单元是一个程序集。

但是还有InternalsVisibleToAttribute(otherAssembly)，它的作用类似于跨组装的友元机制。微软将此用于可视化设计器程序集。

首先(依我看)不要听那些说朋友没用的人。它是有用的。在许多情况下，您将拥有具有不打算公开可用的数据或功能的对象。对于许多作者可能只是表面上熟悉不同领域的大型代码库尤其如此。

友元说明符也有替代方案，但通常都很麻烦(cppp级别的具体类/掩码类型定义)，或者不是万无一错(注释或函数名约定)。

在答案上;

友元说明符允许指定类访问发出友元语句的类内受保护的数据或功能。例如，在下面的代码中，任何人都可以询问孩子的名字，但只有母亲和孩子可以更改名字。

您可以通过考虑一个更复杂的类(如Window)来进一步考虑这个简单的示例。一个窗口很可能会有许多不应该被公开访问的函数/数据元素，但是被相关的类(如WindowManager)所需要。

class Child
{
//Mother class members can access the private parts of class Child.
friend class Mother;

public:

  string name( void );

protected:

  void setName( string newName );
};

In C++ "friend" keyword is useful in Operator overloading and Making Bridge. 1.) Friend keyword in operator overloading :Example for operator overloading is: Let say we have a class "Point" that has two float variable"x"(for x-coordinate) and "y"(for y-coordinate). Now we have to overload "<<"(extraction operator) such that if we call "cout << pointobj" then it will print x and y coordinate (where pointobj is an object of class Point). To do this we have two option: 1.Overload "operator <<()" function in "ostream" class. 2.Overload "operator<<()" function in "Point" class. Now First option is not good because if we need to overload again this operator for some different class then we have to again make change in "ostream" class. That's why second is best option. Now compiler can call "operator <<()" function:

   1.Using ostream object cout.As: cout.operator<<(Pointobj) (form ostream class).
   2.Call without an object.As: operator<<(cout, Pointobj) (from Point class).

Beacause we have implemented overloading in Point class. So to call this function without an object we have to add"friend" keyword because we can call a friend function without an object. Now function declaration will be As: "friend ostream &operator<<(ostream &cout, Point &pointobj);" 2.) Friend keyword in making bridge : Suppose we have to make a function in which we have to access private member of two or more classes ( generally termed as "bridge" ) . How to do this: To access private member of a class it should be member of that class. Now to access private member of other class every class should declare that function as a friend function. For example : Suppose there are two class A and B. A function "funcBridge()" want to access private member of both classes. Then both class should declare "funcBridge()" as: friend return_type funcBridge(A &a_obj, B & b_obj);I think this would help to understand friend keyword.

在工作中，我们广泛地请朋友来测试代码。这意味着我们可以为主应用程序代码提供适当的封装和信息隐藏。但我们也可以有单独的测试代码，使用朋友来检查内部状态和数据进行测试。

可以说，我不会将friend关键字作为设计的重要组成部分。

正如朋友声明的参考资料所说:

友元声明出现在类主体中，并授予函数或另一个类对出现友元声明的类的私有和受保护成员的访问权。

所以提醒一下，有些回答中有技术错误，说朋友只能访问受保护的成员。

这可能不是一个实际的用例情况，但可能有助于说明类间朋友关系的使用。

会所

class ClubHouse {
public:
    friend class VIPMember; // VIP Members Have Full Access To Class
private:
    unsigned nonMembers_;
    unsigned paidMembers_;
    unsigned vipMembers;

    std::vector<Member> members_;
public:
    ClubHouse() : nonMembers_(0), paidMembers_(0), vipMembers(0) {}

    addMember( const Member& member ) { // ...code }   
    void updateMembership( unsigned memberID, Member::MembershipType type ) { // ...code }
    Amenity getAmenity( unsigned memberID ) { // ...code }

protected:
    void joinVIPEvent( unsigned memberID ) { // ...code }

}; // ClubHouse

会员班的

class Member {
public:
    enum MemberShipType {
        NON_MEMBER_PAID_EVENT,   // Single Event Paid (At Door)
        PAID_MEMBERSHIP,         // Monthly - Yearly Subscription
        VIP_MEMBERSHIP,          // Highest Possible Membership
    }; // MemberShipType

protected:
    MemberShipType type_;
    unsigned id_;
    Amenity amenity_;
public:
    Member( unsigned id, MemberShipType type ) : id_(id), type_(type) {}
    virtual ~Member(){}
    unsigned getId() const { return id_; }
    MemberShipType getType() const { return type_; }
    virtual void getAmenityFromClubHouse() = 0       
};

class NonMember : public Member {
public:
   explicit NonMember( unsigned id ) : Member( id, MemberShipType::NON_MEMBER_PAID_EVENT ) {}   

   void getAmenityFromClubHouse() override {
       Amenity = ClubHouse::getAmenity( this->id_ );
    }
};

class PaidMember : public Member {
public:
    explicit PaidMember( unsigned id ) : Member( id, MemberShipType::PAID_MEMBERSHIP ) {}

    void getAmenityFromClubHouse() override {
       Amenity = ClubHouse::getAmenity( this->id_ );
    }
};

class VIPMember : public Member {
public:
    friend class ClubHouse;
public:
    explicit VIPMember( unsigned id ) : Member( id, MemberShipType::VIP_MEMBERSHIP ) {}

    void getAmenityFromClubHouse() override {
       Amenity = ClubHouse::getAmenity( this->id_ );
    }

    void attendVIPEvent() {
        ClubHouse::joinVIPEvent( this->id );
    }
};

设施

class Amenity{};

如果你看看这些类之间的关系;会所拥有各种不同类型的会员资格和会员资格。成员都派生自超类或基类，因为它们都共享公共的ID和枚举类型，外部类可以通过基类中的访问函数访问它们的ID和类型。

然而，通过这种成员及其派生类的层次结构以及它们与ClubHouse类的关系，派生类中唯一具有“特殊特权”的是VIPMember类。基类和其他2个派生类不能访问ClubHouse的joinVIPEvent()方法，但VIP Member类拥有该特权，就好像它拥有对该事件的完全访问一样。

所以对于vip会员和ClubHouse，这是一个双向通道，而其他会员职业是有限的。