我一直在阅读c++常见问题,对朋友声明很好奇。我个人从未使用过它,但我对探索这门语言很感兴趣。

使用friend的一个好例子是什么?


读了一些常见问题,我喜欢<< >>操作符重载和添加作为这些类的朋友的想法。然而,我不确定这如何不会破坏封装。什么时候这些异常可以保持在OOP的严格范围内?


当前回答

对于操作符<<和操作符>>,没有很好的理由让这些操作符成为朋友。它们确实不应该是成员函数,但它们也不需要是朋友函数。

最好的方法是创建公共打印(ostream&)和读取(istream&)函数。然后,根据这些函数写出操作符<<和操作符>>。这提供了额外的好处,允许您将这些函数设置为虚拟函数,从而提供虚拟序列化。

其他回答

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。

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.

在为类实现树算法时,教授给我们的框架代码将树类作为节点类的朋友。

它实际上没有任何好处,除了让你在不使用设置函数的情况下访问成员变量。

在做TDD的时候,我经常使用c++中的'friend'关键字。

朋友能知道我的一切吗?


更新:我从Bjarne Stroustrup网站上找到了这个关于“朋友”关键字的有价值的答案。

“好友”是一种授予访问权限的显式机制,就像会员资格一样。

典型的例子是重载操作符<<。另一个常见的用法是允许助手或管理类访问您的内部。

下面是我从c++朋友那里听到的一些指导原则。最后一个尤其令人难忘。

你的朋友不是你孩子的朋友。 你孩子的朋友并不是你的朋友。 只有朋友才能碰你的隐私部位。