本文介绍了通过具有virtualdtor的Base对象删除派生对象的[]的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

假设


班级基础

{

公开:

虚拟〜测试() {...}

// ...

};


class派生:公共基地

{

public:

virtual~Endived(){...}

// ...

};


int main()

{

Base * base_ptr = new Derived [10]();

delete [] base_ptr;

返回EXIT_SUCCESS;

}


如果是基类dtor不是虚拟的,''删除[] base_ptr''有

未定义的行为。


将''删除[] base_ptr''调用每个Derived类dtor因为

Base :: ~Base()是虚拟的吗?以上代码中的删除是否有效?。

或者这是否也会调用未定义的行为?


请澄清。


谢谢

V.Subramanian

Suppose

class Base
{
public:
virtual ~Test() { ... }
// ...
};

class Derived : public Base
{
public:
virtual ~Derived() { ... }
// ...
};

int main()
{
Base* base_ptr = new Derived[10]();
delete [] base_ptr;
return EXIT_SUCCESS;
}

If the Base class dtor is not not virtual, ''delete [] base_ptr'' has
undefined behaviour.

Will ''delete [] base_ptr'' call each Derived class dtor because the
Base::~Base() is virtual ? Is the deletion in the above code valid ?.
Or does this also invoke undefined behaviour ?

Kindly clarify.

Thanks
V.Subramanian

推荐答案



不,我们不能这么认为。这不是一个有效的C ++类定义。


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No, we can''t suppose that. This is not a valid C++ class definition.

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你的意思是

虚拟~Base(){/*...*/}

这里?

Did you mean
virtual ~Base() { /*...*/ }
here?



如果Base使用正确的虚拟Base析构函数是正确的(而不是测试

析构函数是无效的)我相信这是格式良好的代码。


-

Jim Langston




您的意思是

虚拟~Base(){/*...*/}

在这里?


Did you mean
virtual ~Base() { /*...*/ }
here?



如果Base使用正确的虚拟Base析构函数是正确的(而不是测试

析构函数是无效的)我相信这是格式良好的代码。


If Base is correct with a proper virtual Base destructor (and not Test
destructor which is invalid) I believe this is well formed code.



形成良好 - 也许。但是代码有[5.3.5 / 3]未定义的行为:


...在第二种方法(删除数组)中如果动态类型为

要删除的对象与其静态类型不同,行为是

undefined。

最佳


Kai-Uwe Bux

Well formed -- maybe. But the code has undefined behavior as per [5.3.5/3]:

... In the second alternative (delete array) if the dynamic type of the
object to be deleted differs from its static type, the behavior is
undefined.
Best

Kai-Uwe Bux


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10-27 15:03