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From: | Ludvig Larsson <ludvig AT club-internet DOT fr> |
Newsgroups: | comp.os.msdos.djgpp |
Subject: | Re: superslow simpel rep stosl, why? |
Date: | Mon, 19 Oct 1998 20:17:49 +0200 |
Organization: | Faas-Goldhart |
Lines: | 25 |
Message-ID: | <362B824C.6A9B@club-internet.fr> |
References: | <Pine DOT SUN DOT 3 DOT 91 DOT 981019102612 DOT 7874P-100000 AT is> |
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NNTP-Posting-Date: | 19 Oct 1998 18:30:41 GMT |
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To: | djgpp AT delorie DOT com |
DJ-Gateway: | from newsgroup comp.os.msdos.djgpp |
Reply-To: | djgpp AT delorie DOT com |
Eli Zaretskii wrote: > > On Mon, 19 Oct 1998, Ludvig Larsson wrote: > > > On my AmdK6-2 300mhz it takes 0.006 sec. which gives about > > 100millions of bytes/sec. Quite a bit right! > > But should it take 3 clockcykles to clear each byte? > > I'm clearing quadwords... > > > > I'm using asm(rep stosl). > > > > Is this normal? > > Why not? On a 486 STOSD is documented to require 5 clocks per move, > so it doesn't strike me as terribly wrong to get 3 clocks on K6. Keep > in mind that it doesn't just move the dword, it also increments a > pointer and decrements a count as it goes. > But? As I'm clearing d-words, each stosl takes 12 cycles... I have trought of a way so I don't need to clear the z-buffer, but it should be really nice to push the processor under the "4-times to slow" limit. Ludvig Larsson
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