BUG：编译器允许在没有编译析构函数的情况下创建对象 [英] BUG: compiler allows for creation of objects without destructor compiled
问题描述
以下代码：
#include< iostream>
班级基础
{
私人：
虚拟~base（）
{
std :: cout<< " virtual~base（）\ n";
}
};
派生类：公共基础//必须导致编译时错误
{
//在取消注释时确实会导致编译时错误
//〜derived（）{ }
};
int main（）
{
//确实导致了取消注释时编译时错误
//派生dd;
派生* d =新派生;
删除d;
}
编译时认为必须产生错误。它只产生2
警告：
警告C4624：''衍生''：无法生成析构函数因为
基类析构函数不可访问
警告C4527：类''derived''的实例永远不会被销毁 -
用户定义的析构函数需要
第二个警告是假的 - 代码确实创建了派生的实例
，更糟糕的是它会破坏它。在运行时，它看起来像是基础的'
析构函数没有被调用。
-
Maxim Yegorushkin
MetaCommunications Engineering
http：//www.meta- comm.com/engineering/
The following code:
#include <iostream>
class base
{
private:
virtual ~base()
{
std::cout << "virtual ~base()\n";
}
};
class derived : public base // must result in a compile time error
{
// does result in a compile time error when uncommented
//~derived() {}
};
int main()
{
// does result in a compile time error when uncommented
//derived dd;
derived* d = new derived;
delete d;
}
Produces no error when compiled thought it must. It only produces 2
warnings:
warning C4624: ''derived'' : destructor could not be generated because a
base class destructor is inaccessible
warning C4527: instances of class ''derived'' can never be destroyed -
user-defined destructor required
The second warning is false - the code does create an instance of derived
and what much worse it does destroy it. At runtime it seems like base''s
destructor does not get called.
--
Maxim Yegorushkin
MetaCommunications Engineering
http://www.meta-comm.com/engineering/
推荐答案
您好，
试试这个：
class bas
public
virtual~base（
TRACE（" virtual~base（） \ n"）
}
派生类：公共基础//必须导致编译时间错误
//在uncommente时确实会导致编译时错误
public：
~derived（）{}
}
查看您编写的代码。：您已经将析构函数设置为私有。因此它也不会在派生中显示。
这就是为什么它会给编译时错误。
Hi ,
Try this :
class bas
public
virtual ~base(
TRACE("virtual ~base()\n")
}
class derived : public base // must result in a compile time erro
// does result in a compile time error when uncommente
public :
~derived() {}
}
See the code you have written. : you have made destructor "Private ". So that it would not be visible in derived also.
That''s why it is giving compile time error.
Rudresh < RU *********** @ efi.com>写道：
Rudresh <ru***********@efi.com> wrote:
[...]
查看您编写的代码。 ：你已经使析构函数私有。
因此它也不会在派生中可见。
这就是它给编译时错误的原因。
[...]
See the code you have written. : you have made destructor "Private ".
So that it would not be visible in derived also.
That''s why it is giving compile time error.
再次查看帖子。我会说：
这就是为什么它_should_给出了一个
编译时错误。
我我称这是一个错误。 Comeau也是如此：
Comeau C / C ++ 4。3。3（2003年8月6日15:13:37）ONLINE_EVALUATION_BETA1
版权所有1988-2003 Comeau Computing。保留所有权利。
模式：严格错误C ++
" ComeauTest.c"，第12行：错误：" base :: ~base（） "无法访问
类派生：公共基础//必须导致编译时错误
^
在隐式生成derived时检测到： ：衍生（）"在线
23
" ComeauTest.c"，第12行：错误：" base :: ~base（）"无法访问
类派生：公共基础//必须导致编译时错误
^
期间检测到：
隐式生成derived ::〜derived（）第23行
隐式生成derived :: derived（）第23行
在编辑ComeauTest.c中检测到2个错误。
Schobi
-
Sp******@gmx.de 永远不会被阅读
我是Schobi at suespammers dot org
有时编译器比人们更合理。
Scott Meyers
Look at the posting again. I''d say:
"That''s why it _should_ be giving a
compile-time error."
I''d call this a bug. So does Comeau:
Comeau C/C++ 4.3.3 (Aug 6 2003 15:13:37) for ONLINE_EVALUATION_BETA1
Copyright 1988-2003 Comeau Computing. All rights reserved.
MODE:strict errors C++
"ComeauTest.c", line 12: error: "base::~base()" is inaccessible
class derived : public base // must result in a compile time error
^
detected during implicit generation of "derived::derived()" at line
23
"ComeauTest.c", line 12: error: "base::~base()" is inaccessible
class derived : public base // must result in a compile time error
^
detected during:
implicit generation of "derived::~derived()" at line 23
implicit generation of "derived::derived()" at line 23
2 errors detected in the compilation of "ComeauTest.c".
Schobi
--
Sp******@gmx.de is never read
I''m Schobi at suespammers dot org
"Sometimes compilers are so much more reasonable than people."
Scott Meyers
Rudresh写道：
Rudresh wrote:
查看您编写的代码。 ：你已经使析构函数私有。所以
这是我的意图。
它也不会在派生中看到。
这就是它给编译时间的原因错误。
See the code you have written. : you have made destructor "Private ". So
That is my intent.
that it would not be visible in derived also.
That''s why it is giving compile time error.
问题是它*不会*给出编译错误但必须。
-
Maxim Yegorushkin
The problem is that it *does not* give a compile error but it must.
--
Maxim Yegorushkin
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