Message-ID:	<329E319E.2A82@gbrmpa.gov.au>
Date:	Fri, 29 Nov 1996 08:43:11 +0800
From:	Leath Muller <leathm AT gbrmpa DOT gov DOT au>
Reply-To:	leathm AT gbrmpa DOT gov DOT au
Organization:	Great Barrier Reef Marine Park Authority
MIME-Version:	1.0
To:	Glen Miner <gminer AT Newbridge DOT COM>
CC:	djgpp AT delorie DOT com
Subject:	Re: Optimization
References:	<57hg9b$or5 AT kannews DOT ca DOT newbridge DOT com> <329C95AD DOT C3E AT silo DOT csci DOT unt DOT edu> <57k531$5bu AT kannews DOT ca DOT newbridge DOT com>

> > There shouldnt be any problem running 16 bit data in 32 bit form.
> > I don't understand what is your problem here??? How did you get
> > the feeling that your program would actually get slower if you use
> > 32 bit data instead of 16 bit data.
> 
> Well, my logic is this: You have to move 2x as much data around; this
> means your L1 cache fills up 2x as fast. This is not good.

Hmmm...I would think that if about 20 people told me one thing, and I
was the only person considering the other, that I would be wrong... ;)

Anyway, _regardless_ of size of addressed data on the pentium, it always
loads in 32 bytes into its cache. If you access a byte, short int or
int,
the cache will _always_ be filled with 32 bytes of memory.

If you are loading 16 bit values, you can get effectively 16 pieces of
data into the 32 byte cache line, whereas you only get 8 pieces of 32
bit data.

Now, here's the difference: if your accessing 32 bit data, and its in
the
cache, you get it for free (ie: 0 cycles), if you access 16 bit data,
you have a size prefix (1 cycle).

So, to access the 16 pieces of 16 bit code, it would take 16 cycles.

To access 16 pieces of 32 bit code, it would take 1 cycle. Why? Because
a movl from memory takes 1 cycle. If you have 64 bytes of data, you will
do one extra cache load, which takes a cycle. (Or maybe 2, its still
better than 16!!!)

Now, did you follow this ok???

Leathal.

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