Path: utzoo!utgpu!water!watmath!clyde!rutgers!bellcore!decvax!decwrl!pyramid!oliveb!sun!pepper!cmcmanis
From: cmcmanis%pepper@Sun.COM (Chuck McManis)
Newsgroups: comp.sys.amiga.tech
Subject: ALERT, bug in DOSAlloc code
Keywords: BUG, example code, ECO
Message-ID: <46937@sun.uucp>
Date: 24 Mar 88 23:57:30 GMT
Sender: news@sun.uucp
Lines: 47

Bob Page pointed out to me that the formula I was using to calculate
allocation lengths in my DOSAlloc() example were broken. I checked on
my original source and found out where I screwed up. As mentioned 
previously, DOS allocate things in integral numbers of long words and
stores the length in the long word at the -1 offset. The generic DOS alloc
looks something like :

			  +------------+
		  	  |   length   |
Returned ptr (long *) --->+------------+
			  | long 0     |
			  +------------+
				...
			  +------------+
		          | long n     |
			  +------------+

So if you are allocating a BPTR that DOS may want to unallocate sometime
you can calculate the memory you need with the formula
	((((size+3)/4)+1)*sizeof(long))

Which figures out how many longwords you need (size+3)/4  [ size is in BYTES ]
adds one to it for the length longword, and then multiplies by the size
of a long word to get a value back into BYTES. This is passed to AllocMem(). 
This same number is stored in the length, so the new DOSAlloc() looks like:

long	*
DOSAlloc(size)

int	size;
{
	ULONG *T; 
	T = (long *)AllocMem((((size+3)/4)+1)*sizeof(long));
	if (!T) return(NULL);
	*T = ((((size+3)/4)+1)*sizeof(long));
	return(T+1);
}

DOSFree can still be defined as 
#define DOSFree(p)	FreeMem((long *)p -1,*((long *)p - 1))

Sorry for the junk earlier.

--Chuck
--Chuck McManis
uucp: {anywhere}!sun!cmcmanis   BIX: cmcmanis  ARPAnet: cmcmanis@sun.com
These opinions are my own and no one elses, but you knew that didn't you.