Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: Notesfiles $Revision: 1.7.0.8 $; site uiucdcsb Path: utzoo!watmath!clyde!cbosgd!ihnp4!inuxc!pur-ee!uiucdcsb!guzzi From: guzzi@uiucdcsb.CS.UIUC.EDU Newsgroups: net.micro Subject: Re: AT&T PC6300 PLUS 1meg on motherboar Message-ID: <4400124@uiucdcsb> Date: Wed, 16-Oct-85 11:44:00 EDT Article-I.D.: uiucdcsb.4400124 Posted: Wed Oct 16 11:44:00 1985 Date-Received: Sat, 19-Oct-85 05:14:46 EDT References: <352@cbuxc.UUCP> Lines: 29 Nf-ID: #R:cbuxc.UUCP:-35200:uiucdcsb:4400124:000:1188 Nf-From: uiucdcsb.CS.UIUC.EDU!guzzi Oct 16 10:44:00 1985 I think that MS-DOS has a 640k limit on the amount of memory that it will address, which may explain why the diagnostics report only 512k. But, I am a bit confused. I know that the 8086 processor in the pc6300 has 20 address bits == 1meg, but what about the video ram? I have a pc6300 with 640k in it. The norton utilitites report the following: 640k from hex paragraph 0000 to A000 (a hex paragraph is 16 bytes) 32k from hex paragraph B800 to C000 (some of this may be phantom memory) I assumed that the extra 32k was the 32k of video ram. Hex paragraph B800 is address 736k, which is right in the middle of your 1 meg of memory. Is this 32k real memory? If not, then can you really address all 1 meg of ram using either a ram disk or another operating system, like XENIX? --Mark Guzzi University of Illinois at Urbana-Champaign ARPA: guzzi@uicsrd.csrd.uiuc.edu guzzi%uicsrd.csrd@uiuc.edu guzzi%uicsrd@uiuc.arpa CSNET: guzzi%uicsrd@uiuc guzzi%uicsrd.csrd.uiuc.edu@csnet-relay UUCP: {ihnp4,convex,pur-ee}!uiucdcs!guzzi USENET: ...!{pur-ee,ihnp4}!uiucdcs!guzzi BITNET: guzzi%uicsrd.csrd.uiuc.edu@WISCVM.BITN