Path: utzoo!attcan!uunet!van-bc! From: lphillips@lpami.wimsey.bc.ca (Larry Phillips) Newsgroups: comp.sys.amiga Subject: Re: A2630 board Message-ID: <1204@lpami.wimsey.bc.ca> Date: 5 Mar 90 22:19:11 GMT Lines: 35 Return-Path: To: van-bc!rnews In <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? > A byte is 8 bits. A megabyte is 1048576 bytes (1K * 1K, or 1024 * 1024 bytes). The 'things' that you are speaking of are not bytes. 16 chips, in the 256*4 variety, can hold 2 megabytes of data, and can be organized as: 2 megabytes (2097152 by 8 bits wide) 1 megawords (1048576 by 16 bits wide) .5 mega-longwords ( 524288 by 32 bits wide) There are, of course, other ways to organize it, but these are the ways you will usually see it laid out. The first organization is pretty much useless on an Amiga, unless it's on a peripheral card that needs it that way for some reason. Note, though, that in all cases, we have the same number of bytes. -larry -- Gallium Arsenide is the technology of the future; always has been, always will be. +-----------------------------------------------------------------------+ | // Larry Phillips | | \X/ lphillips@lpami.wimsey.bc.ca -or- uunet!van-bc!lpami!lphillips | | COMPUSERVE: 76703,4322 -or- 76703.4322@compuserve.com | +-----------------------------------------------------------------------+