Re: A89: Shifting/Multiplying. . .


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Re: A89: Shifting/Multiplying. . .




I think that mulu would be a bit faster.  Give or take a byte, each
instruction is three bytes long.  Seeing that the processor is 32-bit, it
doesn't really matter in performance speed, but if you have many
instances of it, go with the mulu.  You'll find that your program will
run faster, since every time you want to call the subroutine, you need to
tack on a few more bytes (to save program counter, push arguments, etc.).
 But I'm just a beginner.  Zoltan, insert your words of wisdom. =) 

---------------------------------------------------------------------------------
Cody Maggard, Technomancer
RavenLoft_15@juno.com

On Mon, 13 Sep 1999 19:06:03 -0400 "Scott Noveck" <noveck@pluto.njcc.com>
writes:
>
>Another optimization (note: "optomization" is spelled with a _z_ =)
>question:  If I'm multiplying one variable number by a constant that 
>is the
>sum of several powers of two, is it better to flat-out multiply or to
>perform several left shifts (multiplying by the powers of two) and 
>adding
>those numbers?
>
>For example:  I want to multiply an arbitrary variable value contained 
>in d0
>by 12.  Ten is the sum of 2 powers of two: 8 (2^3) and 4 (2^2).  I can 
>use a
>simple "mulu.w #12,d0" or I can use the following little algorithm:
>
>mul_d0_12:
> move.w d0,d1    ;arbitrary value is a word
> lsl.w #2,d0     ;d0 = (4 * d0)
> lsl.w #3,d1     ;d1 = (8 * d0)
> add.w d1,d0     ;d0 = ((8 * d0) + (4 * d0))
>                 ;   = (d0 * (8 + 4))
>                 ;   = (d0 * 12)
>                 ;[via distributive property of multiplication]
>
>I'm pretty sure that the algorith is a good deal faster and not too 
>much
>larger, although Motorola never did send me my manual so I could tally 
>this
>stuff up myself (Olle, how do I request it again?).  If somone could 
>just
>give me a count of the size/speed of the algorith compared to the
>multiplication (or even show me a better way to do it -- Zoltan =) I'd
>appreciate it.
>
>Also, to what point is this more efficient?  If I were to expand upon 
>this
>to multiply by 31 -- 2^4 + 2^3 + 2^2 + 2^1 + 2^0 (w/ 2^0 not requiring 
>any
>shifting), would it still be more efficient than multiplying?  It's
>beginning to seem to me like the 68k's multiplication/division 
>operations
>are more or less wasteful. . .
>
>    -Scott
>
>

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