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From: andrewt@watnow.waterloo.edu (Andrew Thomas)
Newsgroups: comp.sys.amiga.tech
Subject: Re: Manx-C and realloc()
Message-ID: <ANDREWT.90Apr10120024@watnow.waterloo.edu>
Date: 10 Apr 90 16:00:24 GMT
References: <36252@dcatla.UUCP> <440@oregon.oacis.org>
	<3418@newton.physics.purdue.edu> <28705@cup.portal.com>
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In-reply-to: Sullivan@cup.portal.com's message of 7 Apr 90 21:52:35 GMT

In article <28705@cup.portal.com> Sullivan@cup.portal.com (sullivan - segall) writes:
> >I am using Manx 3.6a, and had the need for realloc() the other
> >day.  Unfortunately the linker couldn't find it, and neither could I!
> 
> 	#define realloc(x,s) (free(x),malloc(x,s))

Doesn't realloc also copy the contents of the original memory, so you
would need something like:

void* realloc (void* x, int oldsize, int size)
{
  void*	y;

  if (size == 0) return (NULL);
  y = (void*) malloc (size);
  if (y == NULL) return (NULL);
  if (oldsize > size) oldsize = size;
  bcopy (x, y, oldsize);
  free (x);
  return (y);
}

Realloc guarantees that the contents don't change from the old chunk
of memory to the new one, up to the smaller of the two block sizes.
Without knowing the format of the memory list header, you cannot find
the size of the old chunk, so you have to send it to the homebrew
realloc yourself.  This is why realloc is supposed to be part of the
standard memory management library.

The above is UNIX code, not Amiga.  I have never encountered a
two-argument malloc command (shows how much Amiga programming I do).
In order for the preceding #define to work, though, malloc (x,y) would
have to be identical to realloc(x,y) except that it does not free x.
This I doubt.

Hope this helps.
--

Andrew Thomas
andrewt@watnow.waterloo.edu	Systems Design Eng.	University of Waterloo
"If a million people do a stupid thing, it's still a stupid thing." - Opus