Function: Divide 16 bits by 8 bits

[SeaGtGruff]

I've already posted this function in another thread, but I'm reposting it as a separate thread for increased visibility.

 

Here's a user-defined function that you can add to your bB program if you want to be able to divide a 16-bit number by an 8-bit number. It's taken directly from a routine I found here:

 

http://6502org.wikidot.com/software-math-intdiv

 

   function div16by8
  asm
  LDX #8
  ASL temp2
div16by8_1
  ROL
  BCS div16by8_2
  CMP temp3
  BCC div16by8_3
div16by8_2
  SBC temp3
  SEC
div16by8_3
  ROL temp2
  DEX
  BNE div16by8_1
  RTS
end

To call the function, you must pass it three parameters. When the function exits, the two parts of the answer will be in the accumulator and temp2.

 

The following pseudocode example uses the function to divide a 16-bit integer by an 8-bit integer, giving a 16-bit integer as the result:

 

   quotient_hibyte = div16by8 (numerator_hibyte, numerator_lobyte, divisor)
  quotient_lobyte = temp2

This next pseudocode example uses the function to divide an 8.8 fixed-point (16-bit) number by an 8-bit integer, giving an 8.8 fixed-point number as the result:

 

   quotient_wholepart = div16by8 (numerator_wholepart, numerator_fractionalpart, divisor)
  quotient_fractionalpart = temp2

As you can see, they're exactly the same, except for the way we interpret the parameters and results.

 

This routine is for *unsigned* values, and hasn't been tested with "negative" numbers.

 

Michael

[SeaGtGruff]

Please disregard my previous explanation. This routine does *not* work as advertised-- although it turns out that it *does* do exactly what I was originally looking for. I need to revise it a little bit, then post a new explanation.

 

Michael

Edited January 28, 2008 by SeaGtGruff