Re: operator new (iNumBytes) vs std::malloc(iNumBytes)

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Mon, 11 May 2009 01:27:19 -0700 (PDT)
Message-ID:
<3807e017-0795-4efa-ad3f-5304d6f35037@r3g2000vbp.googlegroups.com>
On May 10, 11:15 pm, Gerhard Fiedler <geli...@gmail.com> wrote:

James Kanze wrote:

On May 10, 2:12 pm, Andrew Tomazos <and...@tomazos.com> wrote:

Given:

   void f(void* p) { ... }

and assuming the operator new has not been overloaded, what
are the concrete differences in behaviour of:

    void* p = operator new(iNumBytes);
    f(p);

vs

    void* p = std:malloc(iNumBytes);
    f(p);

That are visible by f?


The main one is that in the first, p must be freed by ::operator
delete, and in the second, by std::free.


But AFAIK that's not visible by f, right? If f were to
de/reallocate the memory, there has to be a convention about
how the memory was allocated, as this is not visible by f.


I guess it depends on what you mean by "visible by f". Given
just a pointer, there's no way f can know if it was allocated by
malloc or by new. Or for that matter, whether it was
dynamically allocated, or points to a local variable. On the
other hand, if p was allocated by operator new, and f attempts
to free it with ::free, there will be undefined behavior. Which
means that the results may be very visible (or invisible---you
just don't know).

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