防止在SmartPointer上删除 [英] Preventing delete on a SmartPointer
问题描述
我在Bill Hubauer（*）提出的实施后实施了一个SmartPointer课程。但我也覆盖了运算符*（）
模板< class ObjectType>
类SmartPointer
{
public：
operator ObjectType *（）const
{return Pointer; }
....
私人：
ObjectType *指针;
};
现在我的编译器可以编译：
SmartPointer< Object> s =新对象;
删除s;
有没有办法防止这种情况？
谢谢，
Mathieu
（*）
http://groups.google.com/group/comp....3ddc38a827a930
完整源代码位于：
http://svn.sourceforge.net/viewcvs.c...ter.h?view=log
Hi,
I have implemented a SmartPointer class following the implementation
proposed by Bill Hubauer(*). But I also override the operator * ()
template<class ObjectType>
class SmartPointer
{
public:
operator ObjectType * () const
{ return Pointer; }
....
private:
ObjectType* Pointer;
};
Now my compiler can compile:
SmartPointer<Object> s = new Object;
delete s;
Is there a way to prevent that ?
Thanks,
Mathieu
(*)
http://groups.google.com/group/comp....3ddc38a827a930
Full source code is at:
http://svn.sourceforge.net/viewcvs.c...ter.h?view=log
推荐答案
在文章< 11 ********************** @ h76g2000cwa.googlegroups .com> ;,
" ;马修" <毫安*************** @ gmail.com>写道：
In article <11**********************@h76g2000cwa.googlegroups .com>,
"mathieu" <ma***************@gmail.com> wrote:
我在Bill Hubauer（*）提出的实施后实施了一个SmartPointer课程。但我也覆盖了运算符*（）
模板<类ObjectType>
类SmartPointer
公共：
运算符ObjectType *（）const
{返回指针; }
...
私有：
ObjectType *指针;
};
现在我的编译器可以编译：
智能指针<对象> s = new Object;
删除s;
有没有办法防止这种情况？
Hi,
I have implemented a SmartPointer class following the implementation
proposed by Bill Hubauer(*). But I also override the operator * ()
template<class ObjectType>
class SmartPointer
{
public:
operator ObjectType * () const
{ return Pointer; }
...
private:
ObjectType* Pointer;
};
Now my compiler can compile:
SmartPointer<Object> s = new Object;
delete s;
Is there a way to prevent that ?
是的。如果你没有明确地告诉编译器这是允许的，那么它就不会允许它。这意味着删除运算符ObjectType *（）。
Yes. If you don''t explicitly tell the compiler that this is allowed, it
won''t allow it. That means remove the operator ObjectType*().
Daniel T.写道：
Daniel T. wrote:
智能指针<对象> s = new Object;
删除s;
有没有办法防止这种情况？
SmartPointer<Object> s = new Object;
delete s;
Is there a way to prevent that ?
是的。如果你没有明确地告诉编译器这是允许的，它就不会允许它。这意味着删除运算符ObjectType *（）。
Yes. If you don''t explicitly tell the compiler that this is allowed, it
won''t allow it. That means remove the operator ObjectType*().
当然，修复是最好的风格。 （这就是为什么C ++标准库如果没有转换运算符那么几乎没有
。）
但是模板化的类也不能提供运算符删除？然后它可以正确地zilch它拥有指针。
-
Phlip
http://c2.com/cgi/wiki?ZeekLand < - 不是博客!! ！
Of course that fix is the best style. (It''s why the C++ Standard Library has
few if no conversion operators.)
But can''t the templated class also provide an operator delete? Then it could
correctly zilch its owning pointer.
--
Phlip
http://c2.com/cgi/wiki?ZeekLand <-- NOT a blog!!!
Daniel T.写道：
Daniel T. wrote:
是的。如果你没有明确地告诉编译器这是允许的，它就不会允许它。这意味着删除运算符ObjectType *（）。
Yes. If you don''t explicitly tell the compiler that this is allowed, it
won''t allow it. That means remove the operator ObjectType*().
那么还有 - 至少 - 在
中定义~Object的另一种方式受保护的部分，但这样做会阻止我在堆栈上定义一个对象。
我以为可能有另一种方式...
-M
Well there is -at least- another way like defining the ~Object in the
protected section, but doing so would prevent me to define an Object on
the stack.
I thought there could be another way...
-M
这篇关于防止在SmartPointer上删除的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
