在多重继承中使用operator delete时,谁调用类的析构函数

Question

这个问题可能听起来太傻了,但是,我没有找到具体的答案.

对于后期绑定的工作原理以及继承中使用的虚拟关键字知之甚少.

与代码示例一样,在继承的情况下,指向在heap和delete运算符上创建的派生类对象的基类指针用于释放内存,派生和基类的析构函数将仅按顺序调用当基础析构函数声明为虚函数时.

现在我的问题是:

1)当base的析构函数不是虚拟的时,为什么只有在使用"delete"运算符的情况下才会出现不调用派生dtor的问题,为什么不在下面给出的情况下:

derived drvd;
base *bPtr;
bPtr = &drvd; //DTOR called in proper order when goes out of scope.

2) When "delete" operator is used, who is reponsible to call the destructor of the class? The operator delete will have an implementation to call the DTOR ? or complier writes some extra stuff ? If the operator has the implementation then how does it looks like , [I need sample code how this would have been implemented].

3) If virtual keyword is used in this example, how does operator delete now know which DTOR to call?

Fundamentaly i want to know who calls the dtor of the class when delete is used.

<h1> Sample Code </h1>

class base
{
    public:
    base(){
       cout<<"Base CTOR called"<<endl;
    }

    virtual ~base(){  
       cout<<"Base DTOR called"<<endl;
    }
};

class derived:public base
{
    public:
        derived(){
            cout<<"Derived CTOR called"<<endl;
        }

    ~derived(){
            cout<<"Derived DTOR called"<<endl;
     }
};

I'm not sure if this is a duplicate, I couldn't find in search.

int main() { base *bPtr = new derived();

delete bPtr;// only when you explicitly try to delete an object
return 0;

}

2) When "delete" operator is used, who is reponsible to call the destructor of the class? The operator delete will have an implementation to call the DTOR ? or complier writes some extra stuff ? If the operator has the implementation then how does it looks like , [I need sample code how this would have been implemented].

3) If virtual keyword is used in this example, how does operator delete now know which DTOR to call?

Fundamentaly i want to know who calls the dtor of the class when delete is used.

derived drvd;
base *bPtr;
bPtr = &drvd; //DTOR called in proper order when goes out of scope.

I'm not sure if this is a duplicate, I couldn't find in search.

int main() { base *bPtr = new derived();

<h1> Sample Code </h1>

class base
{
    public:
    base(){
       cout<<"Base CTOR called"<<endl;
    }

    virtual ~base(){  
       cout<<"Base DTOR called"<<endl;
    }
};

class derived:public base
{
    public:
        derived(){
            cout<<"Derived CTOR called"<<endl;
        }

    ~derived(){
            cout<<"Derived DTOR called"<<endl;
     }
};

}

Answer 1

编译器生成所有必要的代码以正确的顺序调用析构函数，无论是超出范围的堆栈对象或成员变量，还是被删除的堆对象。