Re: Deleting an Object and setting the Pointer to NULL

From:
Seungbeom Kim <musiphil@bawi.org>
Newsgroups:
comp.lang.c++.moderated
Date:
Sat, 17 Mar 2007 14:50:29 CST
Message-ID:
<ethc4a$pfa$1@news.Stanford.EDU>
James Kanze wrote:

There are several very important reasons:

  -- The argument for delete is not required to be an lvalue---in
     fact, in my code, it often isn't. And you can't set a
     non-lvalue to NULL.

  -- It doesn't buy you anything, anyway, since typically,
     there's not just a single pointer to the object, but a
     number of them.


Those are the essential reasons, I believe.

  -- Finally, most of the time when there is just a single
     pointer, it is in an object, and the delete is in the
     destructor, so there's no point---the pointer itself will
     cease to exist.


Except when the pointer is in a container, in which case you have to
explicitly set the pointer to null to avoid undefined behaviour, IIRC:

     for (i = begin(); i != end(); ++i) delete *i, *i = 0;
or
     for (i = begin(); i != end(); ++i) delete i->second, i->second = 0;

In the past I've written the following to avoid repeating the same code:

template <typename T>
inline void delete_ptr(const T*& p)
{ boost::checked_delete(p); p = 0; }

template <typename T>
struct deleter : public std::unary_function<const T*&, void>
{
     void operator()(const T*& p) const { delete_ptr(p); }
};

// a function object that deletes the 'second' members of std::pair's
// using member template - no need to specify the template argument
// when using, but cannot derive from std::unary_function

struct mapped_ptr_deleter
{
     template <typename T1, typename T2>
     void operator()(std::pair<T1, T2>& p) const { delete_ptr(p.second);}
};

--
Seungbeom Kim

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