Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!uunet!drivax!frotz From: frotz@dri.com (Frotz) Newsgroups: comp.sys.ibm.pc.misc Subject: Re: DR-DOS ver 5.0 (now *very* long, biased opinion;-) Message-ID: Date: 22 Aug 90 17:47:40 GMT References: <9K1N7K1@drivax.UUCP> <35010004@hpfinote.HP.COM> <1990Aug21.195452.23710@agate.berkeley.edu> Sender: frotz@dri.com Reply-To: frotz@dri.com Organization: Digital Research, Monterey CA Lines: 63 joonsong@monsoon.Berkeley.EDU (Suk-Hyun Song) writes: ]In article <35010004@hpfinote.HP.COM> pnl@hpfinote.HP.COM (Peter Lim) writes: ]>... you'll never be able to get > 600 K free with MS-DOS ]>and one of QRAM, 386MAX etc. (unless you use hercules card). This is ]>because MS-DOS must reside in low memory. ]I suppose this is all a matter of definition. ]Does a 640K computer have 640000 bytes or 655360? ]If you think 1K = 1024 bytes, then 600K = 614400 bytes. ] Or if 1K = 1000 bytes, then 600K = 600000 bytes. I guess you answered your own question here. 655360. ]Nobody is really wrong. The question is, does DR-DOS 5.0 provide ]620*1000 bytes or 620*1024 bytes? Yes, somebody is wrong. When dealing with computers in general, k==1024. When dealing with mathematics in general, k==1000. When dealing with a mixture of both, it is the responsibility of the speaker to denote what 'k' means. In this case, we are talking computers. Any other use of 'k' is wrong! If you are talking to an end-user (someone not on the net and barely computer literate by our standards) then K will probably be the mathematical definition. I have a 386 Silicon Valley Computer, 4M memory, 104M disk with 30 buffer, 80 files, EMS, ANSI driver, CACHE (1M), Bus Mouse driver, VGA, History v1.41, 26 drive letters and a 3000 byte environment (SHELL=...), COUNTRY.SYS. MAPMEM 2.9 from Turbopower shows 619440 bytes free conventional memory. It also shows the expanded memory usage: block bytes (Expanded Memory) ----- ------ 1 1048576 free 1949696 total 3145628 DRDOS 5.0 MEM shows 655,360 bytes, ( 640K), conventional memory 619,440 bytes, ( 604K), largest available block 3,145,728 bytes, (3072K), extended memory 3,145,728 bytes, (3072K), extended memory used 0 bytes, ( 0K), extended memory available DRDOS 5.0 CACHE shows 1 drive cached in 62 track buffers 1023K bytes of Expanded Memory used 100% of CACHE in use with a 86% hit rate. Now I think that this qualifies as being greater than 600K (K==1024), and admittedly barely, but I do have a lot going on in my environment too. advAPPOLOGIESance if this is inflamatory. -- Frotz Disclaimer: I do not speak for this company this time, as my remarks are potentially inflamatory.