fake 32 bit division for dummies

Lloyds · 2011-03-26 18:51

OK folks,
Please don't laugh and point fingers because I know this is something I should be able to handle myself. But I am stumped.

I have a constant

constantA CON 598760 'which is larger than 16 bits. (the 1's place is a zero and is insignificant)

I want to divide it with a variable

variableB VAR word ' that will range between 370 and 2000

result = constantA / variableB ' the result will range from about 300 to 1500

How do I handle this to get an integer result accurate to the 1's place? Knowing what the tenths place remainder is would be nice, but is not really necessary.

Thanks very much,
Lloyd

Phil Pilgrim (PhiPi) · 2011-03-26 19:03

By "accurate", do you mean 4/5 rounded instead of truncated? If so, do this:

result := (constant + variable >> 1) / variable

-Phil

Lloyds · 2011-03-26 20:21

Hi Phil,
I don't seem to be able to get that to work. Was the colon before the = a typo? I get a syntax error.
Also, with the constant being larger than 16 bits, I get a syntax error of the constant being too large.
I am still stumped. What do you think?
Thanks,
Lloyd

Phil Pilgrim (PhiPi) · 2011-03-26 23:50

Oh, sorry. I thought this was a Propeller question. I didn't notice which forum it was in!

-Phil

Mike G · 2011-03-27 12:29

Not the most sophisticated but it's close. Probably should add modulus logic to reduce the margin of error.

598760 = 0x922E8

' {$STAMP BS2p}
' {$PBASIC 2.5}

lowWord   VAR   Word
highWord  VAR   Word
result    VAR   Word
denom     VAR   Word


Init:
  lowWord = $22E8
  highWord = $9


Main:

  denom  = 370
  GOSUB DoMath
  GOSUB ShowResult
  PAUSE 1000

  denom  = 2000
  GOSUB DoMath
  GOSUB ShowResult
  PAUSE 1000

GOTO Main


DoMath:
  result = ($FFFF / denom) * highWord
  result = result + (lowWord / denom)
RETURN


ShowResult:
  DEBUG CLS, ?result
RETURN

Tracy Allen · 2011-03-27 14:00

' result = 598760 / B
result = (59876 / B * 10) + ((59876 // B * 10 + 0) /

' the last +0 is the 1s digit of the constant, which happens to be zero here.

Here are a couple of links that might help:

http://emesystems.com/BS2math2.htm#Reciprocal shortcut#2

http://emesystems.com/BS2math6.htm#Method3 double prec division 63 bits/31 bits

Lloyds · 2011-03-27 14:46

Mike G,
Cool! I gave it a try and it worked fine but truncated the remainder of the highword division. Came to me after a while.
I added a step to add the remainder from the highword division to the lowword before doint that division. Then it worked great.

DoMath:
result = ($FFFF / denom) * highWord
remainder = $FFFF - (result * denom)
result = result + ((lowWord + remainder) / denom)
RETURN

There is probably a better way to handle the remainder, but it works.
Thanks very much!
Lloyd

Lloyds · 2011-03-27 14:51

Tracy,
Yes, I see what you are doing. I was thinking of trying something like that but was coming up with something much more complicated.
Seems like so much of this 16 bit integer math is learning the tricks and then applying them. I am still having some trouble grasping/visualizing some of the math, but progress is happening.

Thanks everyone,
Lloyd

Tracy Allen · 2011-03-27 20:47

Lloyd,
You mentioned that you might want to carry out the division to the tenths place. The decimal long division can be placed in a loop, as in the following example:

Q VAR WORD   ' quotient
B VAR WORD   ' divisor
D VAR WORD   ' dividend 
n VAR NIB

  D = 59876   ' start with this, most significant digits of 598760
  B = 370
fixedPointDecimalDivide:
  Q = 0
   FOR n=0 TO 2
    Q = Q * 10 + (D / B)
    D = D // B * 10 + 0          ' the + 0 would be needed if there other digits.
  NEXT
  IF D / B > 4 THEN Q = Q + 1  ' round off
  DEBUG CR, DEC Q
  ' e.g., Q=16183, decimal point implied, 1618.3 = 598760 / 370

fake 32 bit division for dummies

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