Path: utzoo!utgpu!news-server.csri.toronto.edu!clyde.concordia.ca!uunet!wuarchive!texbell!bellcore!flash!sdh
From: sdh@flash.bellcore.com (Stephen D Hawley)
Newsgroups: comp.sys.mac.programmer
Subject: Re: malloc() and calloc()
Message-ID: <21250@bellcore.bellcore.com>
Date: 23 Mar 90 19:00:50 GMT
References: <3860@sage.cc.purdue.edu>
Sender: news@bellcore.bellcore.com
Reply-To: sdh@flash.UUCP (Stephen D Hawley)
Organization: Bellcore, Morristown, NJ
Lines: 40

In article <3860@sage.cc.purdue.edu> ar4@sage.cc.purdue.edu (Piper Keairnes) writes:
>   Well the last time this discussion was around, I didn't know that the
>heck these two functions meant (or how C worked for that matter). Things have
>changed, but I believe I'm running into the same problem that others did:
>
>   I CAN do the following successfully:
>
>	int  *i;
>
>	i = (int*) malloc(sizeof(int) * 5);
>
>   but if CANNOT do the following:
>
>	int *i;
>
>	i = (int*) malloc(sizeof(int) * 6);
>
>   Notice the difference??? Trying to allocate 5 gives me a valid pointer.
>Trying to allocate 6 gives me 0x000000. the big NULL! Anyways, anyone know
>why this is happening? I'm on a 2.5meg SE (don't believe a lack of memory
>to be the problem).

Handy hints:

	Use NewPtr() to get memory if you're doing Mac-only stuff.  Think
		C will promote shorts to longs on tollbox calls, so you
		don't have to worry about the argument.

	malloc() takes as its argument type size_t which is an unsigned
		long.  sizeof() evaluates to a short, thus unless you
		#include stdlib.h, you will be sending malloc the wrong
		size.  Try instead i = (int *)malloc(sizeof(int) * 6L);
		or i = (int *)malloc((long)(sizeof(int) * 6));

Steve Hawley
sdh@flash.bellcore.com

"I'm sick and tired of being told that ordinary decent people are fed up with
being sick and tired.  I know I'm certainly not, and I'm sick and tired of
being told that I am."