NetNews Offline 2/NetNews Offline Volume 2.iso/news/comp/std/c++/223

Internet Message Format | 1996-08-06 | 3.1 KB

Path: fido.asd.sgi.com!austern
From: [e-mail] (Steve Clamage)
Newsgroups: comp.std.c++
Subject: Re: new T[0] and sizeof(T)
Date: 01 Feb 1996 17:38:53 PST
Organization: Sun Microsystems Inc.
Approved: [e-mail]
Message-ID: <4erovm$[e-mail]>
References: <[e-mail]>
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Originator: [e-mail]

In article [e-mail], [e-mail] (Marek  MAKOWSKI) writes:
>I would like to ask for comments on three easy questions illustrated
>by the following piece of code:
>
>template <class I, class T>
>void mVect<I,T>::resize(I new_size) {
>   T *old = v;
>   v = new T[new_size];    // T *v is a private member of mVect
>   int size_of_elem = sizeof(T);    // <--- question 3
>   //
>   // do something
>   //
>   delete[] old;     // <--- questions 1 & 2
>}
>
>I have the following questions:
>1. Is it absolutely robust and portable to delete[] old, even
>   if a previous call was for new_size == 0 (or if v was allocated
>   by the ctor for the size == 0) ?

If you allocate an object with new[], you delete it with delete[]. In
addition, it is always safe to delete a null pointer.

>    In other words: is it guaranteed that:
>   (v = new T[0]) == 0;

No. The result of 'new' is never a null pointer (unless the allocation fails
and you are using a pre-exception or "nothrow" version of new). A request of
zero size results in a pointer value distinct from all other pointer values
in the program. So if you write
    T* t1 = new T[count];
    T* t2 = new T[count];
and if the allocations succeed, it must be true that
    ( t1 != 0 && t1 != t2 && t2 != 0 )
even if count is zero. The expression "new T[0]" is invalid, but if you
use a non-const expression for the count, it is OK if the value is zero.

Since you have requested zero objects, you cannot dereference the resulting
pointer -- it might not point at usable storage. But since you acquired the
pointer value from "new" you can (and should) pass it to "delete".

>2. Is it correct to assume that no destructor is called by this statement
>    if old was set as: old = new T[0]  ??

As noted above, "new T[0]" isn't valid, but if you ask for zero objects,
no objects are created. When you delete the array, no destructors
should be called. I can believe that a compiler might get some part of
these requirements wrong, since this sort of thing hardly ever happens and
bugs might slip through testing.

>3. Is there any risk involved in using sizeof(T) in this statement ?

If T is a completely-defined type, you can always ask for sizeof(T); its
value is a compile-time constant. I'm afraid I don't understand this
question.

---
Steve Clamage, [e-mail]
---
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