Path: utzoo!utgpu!jarvis.csri.toronto.edu!cs.utexas.edu!swrinde!ucsd!rutgers!cbmvax!peter
From: peter@cbmvax.commodore.com (Peter Cherna)
Newsgroups: comp.sys.amiga
Subject: Re: A2630 board
Message-ID: <10019@cbmvax.commodore.com>
Date: 6 Mar 90 18:23:13 GMT
References: <1745@crash.cts.com>
Reply-To: peter@cbmvax (Peter Cherna)
Organization: Commodore, West Chester, PA
Lines: 26

In article <1745@crash.cts.com> hawk@pnet01.cts.com (John Anderson) writes:
>
>   Dave, since the A2630 board uses 16 256x4 chips to make 2 megabytes of RAM,
>then(32 bit RAM), then isn't that actually 512K of actual RAM.  Since each
>"thing" needs 32 bits then that means it can only hold 512 thousand "Things"
>since each one is 32 bits long?  Is that correct?  And if so, or not so, how
>much RAM will be left over after your SetCPU program copies Kickstart into 32
>bit RAM?  

When people talk about K, by convention they are referring to thousands
of 8-bit-bytes.  So each "K" is 8192 bits regardless of the memory-word size.
So when we talk of two megabytes of RAM, while it is true that is only 524288
32-bit words, it is still has as much storage capacity of 2 Meg of 16-bit
memory, such as on a standard 2 MB expansion card like the A2052, which has
1048576 16-bit words.  Don't forget that your 512 thousand "things" hold twice
as much memory as 16-bit wide things.  When Kickstart is copied into 32-bit
RAM, it take just as much room, namely 256K of your 2 MB.  All that happens is
that in one access, 32 bits of memory can be gotten, instead of 16.

So you still have the rest of your 2MB (1.75MB) left.

     Peter
--
     Peter Cherna, Software Engineer, Commodore-Amiga, Inc.
     {uunet|rutgers}!cbmvax!peter    peter@cbmvax.cbm.commodore.com
My opinions do not necessarily represent the opinions of my employer.