Retrieving 32bit Multiply Result

[Darrel Taylor]

There's no way to do that with DIV32. The Quotient must be less than 65535.

You would need a TRUE 32 bit divide routine.

Or you could use the Floating Point Math routines from melabs.
http://www.melabs.com/resources/fp.htm

Darrel

[kasapo]

hi darrel ,
do you have some example about the usage of floating point numbers.

after i do my multiplication and division how can i use aint variable?
i have to send the aint to my motor controller byte by byte.

thanks

[mytekcontrols]

Hi Darrel,

I like your ideas about getting at DIV32 system variables, but I have a slightly different situation. How do I use a variable for the divisor that wont violate the 15 bit rule?

I tried the following example, but it don't work (I get: Error Bad expression):

divisor var word
dummy var word

dummy = DIV32 divisor

I really need the ability to alter the divisor via program control, and at times it will need to be bigger then an 8-Bit value. Any ideas?

Thanks,

Edit: Opps! my mistake. It does work properly with a word variable as the divisor, just so long as you stay with the 15 bit rule (32767). This is good news.

[Darrel Taylor]

hehe,

DT

[Josuetas]

I just saw this link. it is of great interest to me right now...

I have this question... how do you add two of these Dwords?

how do i know if the add of two words overflows?
i.e. $1FFFE + $2E

my guess is to add the lower words $FFFE+$2E. But how do i know if there is an overflow? so that i can add one to the higher WORD? even worse..

i.e. $EFFF8 + $6FFE4

My guess again: add two lower words... if over flow add two higher words +1, else add two higher words....

I think i can manage myself once i am sure abour the overflow

Thanks,

[Darrel Taylor]

For words and bytes, you can detect an overflow by testing if the result is less than what you added.

Due to trucation of a 16-bit word ...
$FF00 + $0180 = $0080

So the result will always be less than either of the values added if an overflow occurs.

I think i can manage myself once i am sure abour the overflow

I'm sure you could. But that's just not my way ...

;Initialize your hardware first

A         VAR WORD[2]
B         VAR WORD[2]
Result    VAR WORD[2]
OVRFLOW32 VAR BIT

ASM
; ===== Load 32bit Constant into DWORD =============================
MOVE?CD macro C32in, Dout                      ; Max= 4,294,967,296
    MOVE?CB   low C32in, Dout
    MOVE?CB   low (C32in >> 8), Dout + 1
    MOVE?CB   low (C32in >> 16), Dout + 2
    MOVE?CB   low (C32in >> 24), Dout + 3
  endm
ENDASM

@  MOVE?CD  0xEFFF8, _A  ; Load 983,032 into A
@  MOVE?CD  0x6FFE4, _B  ; Load 458,724 into B

GOSUB Add32

LCDOUT  $FE,1,  "A=   ",HEX4 A[1],":",HEX4 A[0]
LCDOUT  $FE,$C0,"B=   ",HEX4 B[1],":",HEX4 B[0]
LCDOUT  $FE,$90,"Res= ",HEX4 Result[1],":",HEX4 Result[0]
LCDOUT  $FE,$D0,"OVRFLOW32 = ", BIN1 OVRFLOW32
stop

; ===== Add 2 32bit variables ======================================
Add32:
    OVRFLOW32 = 0
    Result[0] = A[0] + B[0]
    IF Result[0] < B[0] then Result[1] = 1
    Result[1] = Result[1] + A[1]
    IF Result[1] < A[1] then OVRFLOW32 = 1
    Result[1] = Result[1] + B[1]
    IF Result[1] < B[1] then OVRFLOW32 = 1
RETURN

Which displays ...

A=   000E:FFF8
B=   0006:FFE4
Res= 0015:FFDC
OVRFLOW32 = 0

HTH,

[Josuetas]

Checking for a smaller result is a simple trick, thanks,

right now my code is never suposed to exced the Max= 4,294,967,296
so i am ignoring the OVRFLOW32 = 0.

This is the code that i am using

Bigword   var word [2]
BigwordL  var BigWord(0)
BigwordH  var BigWord(1)
Bigword1  var word [2]
Bigword1L var BigWord(0)
Bigword1H var BigWord(1)


AddTwoBigWords:   'Adds Bigword1 and BigWord2 result in BigWord
   BigWordH = BigWordH + BigWord1H  ;ignoring any overflow
   BigWordL = BigWordL + BigWord1L
   if BigWordL < BigWord1L then
         BigWordH = BigWordH+1          ;ignoring any overflow
   endif
return

Would this be ok? i made some tests and it seems ok but your experience might see anything wrong here.