Re: std::vector help!!

From:

"Yong Hu" <yhu221300@gmail.com>

Newsgroups:

comp.lang.c++

Date:

28 Jun 2006 13:03:51 -0700

Message-ID:

<1151525031.562658.232860@p79g2000cwp.googlegroups.com>

Richard Herring wrote:

In message <OhUng.256726$Fs1.5467@bgtnsc05-news.ops.worldnet.att.net>,
Howard <alicebt@hotmail.com> writes

"Victor Bazarov" <v.Abazarov@comAcast.net> wrote in message
news:e7oo0b$pem$1@news.datemas.de...

vikram_p_nayak@yahoo.com wrote:

linux_bp wrote:

I have an stl vector array which stores the pointers to objects.
To delete the array i am using:

std::vector<*foo> bar;
...
for (vector<*foo>::iterator itr = bar.begin(); itr != bar.end(); )
{
delete itr;
itr = NULL;
}

Shouldnt this be
delete (*itr);
?
I guess you are deleting the objects being referred to by the vector
elements. Maybe I am wrong.

It's really hard to conclude anything (although you're probably right)
since the code presented is not real code. For example, 'vector<*foo>'
is a definite syntax error.

Not to mention the fact that such a loop would loop forever, since itr is
set to NULL on the first iteration,

And that line will probably only compile at all if vector<T>::iterator
happens to be implemented as T*, which is not necessarily the case.

The vector<T>::iterator is a type defined as T* for sure.

This is how the iterator is defined in vector:

template<class _Ty, class _A = allocator<_Ty> >
    class vector {
public:
                ........
    typedef _A::pointer _Tptr;
    typedef _Tptr iterator;
                ........
}

and the following shows how the _A::pointer is defined in allocator<T>:

template<class T>
    class allocator {
    ...............
    typedef T *pointer;
    typedef const T *const_pointer;
    typedef T& reference;
    ....................
    allocator();
    allocator<T>& operator=(const allocator<T>);
    pointer allocate(size_type n, const void *hint);
    void deallocate(pointer p, size_type n);
    void construct(pointer p, const T& val);
    void destroy(pointer p);
    size_type max_size() const;
    };

Yong Hu

is never changed by the loop statement,
and would thus never equal bar.end().

--
Richard Herring