通过具有virtualdtor的Base对象删除派生对象的[] [英] delete [] of Derived objects through Base object which has virtualdtor
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
su ************** @ yahoo.com ，印度写道：
su**************@yahoo.com, India writes:
假设
类基地
{
公开：
虚拟~Test（）{...}
// ...
};
Suppose
class Base
{
public:
virtual ~Test() { ... }
// ...
};
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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su ************** @ yahoo.com 写道：
假设
班级基础
{
public：
virtual~Test（）{。 ..}
Suppose
class Base
{
public:
virtual ~Test() { ... }
你的意思是
虚拟~Base（）{/*...*/}
这里？
Did you mean
virtual ~Base() { /*...*/ }
here?
// ...
};
class派生：公共基地
{
公开：
虚拟〜衍生（）{...}
//。 ..
};
int main（）
{
Base * base_ptr = new派生[10]（）;
delete [] base_ptr;
返回EXIT_SUCCESS;
}
如果基类dtor不是虚拟，''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？。
或者这也会调用未定义的行为吗？
// ...
};
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 ?
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
ta*******@rocketmail.com
Jim Langston写道：
Jim Langston wrote:
su ************** @ yahoo.com 写道：
su**************@yahoo.com wrote:
>假设
类基地
{
公开：
virtual~Test（）{ ...}
>Suppose
class Base
{
public:
virtual ~Test() { ... }
您的意思是
虚拟~Base（）{/*...*/}
在这里？
Did you mean
virtual ~Base() { /*...*/ }
here?
> // ...
};
类派生：public Base
{
public：
virtual~Endived（）{...}
// ...
};
int main（）
{
Base * base_ptr = new Derived [10]（）;
delete [] base_ptr;
返回EXIT_SUCCESS;
}
如果Base类dtor不是虚拟的，''delete [] base_ptr''具有未定义的行为。
将''delete [] base_ptr''调用每个Derived类dtor，因为
Base :: ~Base（）是虚拟的吗？上述代码中的删除是否有效？
或者这也会调用未定义的行为吗？
>// ...
};
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 ?
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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