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From: usenet@aristeia.com (Scott Meyers)
Newsgroups: comp.std.c++
Subject: Re: Why no size_t to ::operator delete?
Date: Sun, 24 Dec 2006 17:43:13 GMT
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Lucas Galfaso wrote:
 > Scott Meyers wrote:
 >> Can someone explain to me why having the "normal" operator delete take a
 >> size_t is impractical?

 > The answer is simple, there is nowhere to store this information, if
 > you need it, you store it. This is just another example of C++ of not
 > to make everybody pays for what just a few need.

One of us is missing something.  At class scope, operator delete may
take a size_t.  If it does, it's provided by the compiler, which
passes the size of the object.  The information is not "stored" in the
sense of a memory location holding the size -- the compiler passes a
constant.  If a pointer to a derived object via a base class pointer
is being deleted and the base destructor is not virtual, the wrong
size gets passed, but that's okay, behavior under those conditions is
undefined.

You can only delete three things: a pointer to a class type, a pointer
to a non-class type, and a void*.  My suggestion was that a global
operator delete taking a size_t could:

- For pointers to class types, get the same value that'd be passed to
   a class-specific operator delete taking a size.
- For a pointer p to a non-class type, sizeof(*p).
- for a void* pointer, zero.

None of these options requires storing anything anywhere.

Scott

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