topic Re: HC08 16bit Subtraction in 8-bit Microcontrollers

HC08 16bit Subtraction

Bloodhound — Sun, 02 Dec 2007 19:12:39 GMT

Hi all,

Can anybody please provide post some assembler code to subtract an IMM 16bit value from a 16bit value in RAM. Given that the SUB #xx command is limited to an 8bit immediate value I'm not sure how to do this.

Is it a matter of using say the SP and H,X to perform the operation?

I need to be able to handle two possibilites:

RAM Var = $8000

IMM Val = $2000

$8000 - $2000 = $6000

The another scenario is where the RAM value is less than the IMM value...

RAM Var = $2000

IMM Val = $3000

$2000 - $3000 = $F000

Thanks,
Ross

Re: HC08 16bit Subtraction

peg — Sun, 02 Dec 2007 19:32:57 GMT

Hi,

You just do it a byte at a time using SUB for the low byte first and then SBC for the higher bytes.
The examples you chose are not good as they don't produce a carry from the lower byte.

if RAMVar = $2000

LDA RAMVar+1
SUB #$00
STA RESULT+1
LDA RAMVar
SBC #$30
STA RESULT

now
RESULT = $F000

OR to better demonstrate carry:

if RAMVar = $2000

LDA RAMVar+1
SUB #$01
STA RESULT+1
LDA RAMVar
SBC #$30
STA RESULT

now
RESULT = $EFFF

Message Edited by peg on 2007-12-02 10:44 PM

Re: HC08 16bit Subtraction

bigmac — Sun, 02 Dec 2007 22:21:57 GMT

Hello Ross,

For assembly programming, the code will often be easier to follow if constants are represented by a label, as a result of using an equate directive. The following code snippet extends the process described by Peg, when a label is used.

IMM_VAL equ 8192 ; Could be binary, decimal or hex representation

lda RAMvar+1

sub #(IMM_VAL%256) ; Low byte value

sta RESULT+1

lda RAMvar

sbc #(IMM_VAL/256) ; High byte value

sta RESULT

This method should work with most assemblers. Some assemblers have directives to select the low byte or high byte of a word value, and these could alternatively be used.

Regards,

Mac

Message Edited by bigmac on 2007-12-03 01:23 AM

Re: HC08 16bit Subtraction

Bloodhound — Mon, 03 Dec 2007 09:58:08 GMT

Thanks guys, that worked well, though, for those reading this in the future, remember to put a CLC command before the first sub....might save some head scratching.

Cheers,

Ross

Re: HC08 16bit Subtraction

peg — Mon, 03 Dec 2007 10:06:36 GMT

Hi Ross,

The CLC is not required!
SUB does not use the carry only sets/clears due to the result
SBC uses the carry in the operation and sets/clears it due to the result.