Question

这种说法真的意思是,“建筑部有公共继承的违约”。为什么以下法典有误,因为我没有公开关键词,而是从c类中得出?

struct c 
{
protected:
    int i;
public:
    c(int ii=0):i(ii){} 
    virtual c *fun();
};

c* c::fun(){
    cout<<"in c";
    return &c();
}

class d : c
{  
public:
    d(){}
    d* fun()
    {
        i = 9;
        cout<<"in d"<< 	 <<i;
        return &d();
    }
};


int main()
{
    c *cc;
    d dd;
    cc = &dd;
    cc->fun();
}

Answer 1

这意味着:

struct c;
struct d : c

等于

struct d : public c

页: 1 扩大<代码>struct:

struct c;
class d : c;

等于

class d : private c;

由于<条码><>/条码>在违约情况下具有私人继承。

And it means that all inherited and not overriden/overloaded/hidden methods from c are private in d.

Answer 2

“建筑部有公共继承,不成事实”意味着:

struct Derived : Base {};

is equivalent to

struct Derived : public Base {};

Classes have private inheritance by default, so when you remove the public from a class inheritance you have the equivalent of

class Derived : private Base {};

在这种私人继承情景中,Derived没有is-a与Base的关系,实质上是has-aBase。因此,您正试图在此尝试:

cc = &dd;

is not allowed.

Answer 3

当你写上<条码>和从某种东西继承而没有注明出处方时,遗产被作为<条码>公开<>。当你写上<条码>并继承某条,但没有具体说明出处方(即便是哪条是<条码>/条码>),继承即被视为私人继承。在你的法典中,你重新做后一项工作,因此继承是私人的,因此发现的错误。

Put another way, to say that struct inheritance is public by default really means that inheritance done when writing a struct is public by default, not that inheriting from a struct is public by default.

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