PIC Microcontoller Radix Math Method

Binary to BCD packed and ASCII, 32 bit to 10 digits

from Ron Kreymborg and Mike Keitz

[ed: rough guess is that about 2200 instructions will be executed]

see also:

• http://www.jennaron.com/pics/bcd2a.html

;******************************************************************
;
; Test program for 32-bit unsigned binary to BCD and BCD to ASCII.
;
;******************************************************************
        title   "32-bit binary to ascii"
        list p=pic16f84,r=dec,n=80,x=off,st=off
        include <p16f84.inc>
        errorlevel -302         ; no bank warnings
	errorlevel -305		; no default dest warnings

#define	number	123456789

loadd	macro
	local	m,n,p
p = number / 16777216
	movlw	p
	movwf	bin
m = (number - p * 16777216) / 65536
	movlw	m
	movwf	bin+1
n = (number - p * 16777216 - m * 65536) / 256
	movlw	n
	movwf	bin+2
	movlw	number - p * 16777216 - m * 65536 - n * 256
	movwf	bin+3
	endm

	CBLOCK	0x0c
	bin:4		; 32-bit binary number (unsigned)
	bcd:10		; 10 BC digits or 10 ascii chars
	pti,pto		; pointers
	ii
	temp
	cnt
	ENDC
	
	loadd			; load test value <number>
	call	b2bcd		; convert to 32-bit binary to 10 bcd
	call	bcd2a		; convert 10 bcd to 10 ascii
	goto	$

;******************************************************************
; Convert the 10 binary coded digits (5 bytes) starting at 
; <bcd> into an ascii string also starting at <bcd>. Original
; bcd digits are lost.

bcd2a	movlw	bcd+9
	movwf	pto		; destination pointer
	movlw	bcd+4
	movwf	pti		; source pointer
	movlw	5		; 5 bytes to process
	movwf	cnt

bcd2a1	movf	pti,w		; get current input pointer
	movwf	fsr
	decf	pti,f		; prepare for next
	movf	indf,w		; get 2 bcds
	movwf	temp		; save for later
	movf	pto,w		; get current output pointer
	movwf	fsr
	decf	pto,f		; prepare for next
	decf	pto,f
	movf	temp,w		; get digits back
	andlw	0x0f		; process lsd
	addlw	"0"
	movwf	indf		; to output
	decf	fsr,f
	swapf	temp,w		; process msd
	andlw	0x0f
	addlw	"0"
	movwf	indf		; to output
	decfsz	cnt		; all digits?
	goto	bcd2a1
	return			; yes


;******************************************************************
; Convert 32-bit binary number at <bin> into a bcd number
; at <bcd>. Uses Mike Keitz's procedure for handling bcd 
; adjust; Modified Microchip AN526 for 32-bits.

b2bcd	movlw	32		; 32-bits
	movwf	ii		; make cycle counter
	clrf	bcd		; clear result area
	clrf	bcd+1
	clrf	bcd+2
	clrf	bcd+3
	clrf	bcd+4
	
b2bcd2	movlw	bcd		; make pointer
	movwf	fsr
	movlw	5
	movwf	cnt

; Mike's routine:

b2bcd3	movlw	0x33		
	addwf	indf,f		; add to both nybbles
	btfsc	indf,3		; test if low result > 7
	andlw	0xf0		; low result >7 so take the 3 out
	btfsc	indf,7		; test if high result > 7
	andlw	0x0f		; high result > 7 so ok
	subwf	indf,f		; any results <= 7, subtract back
	incf	fsr,f		; point to next
	decfsz	cnt
	goto	b2bcd3
	
	rlf	bin+3,f		; get another bit
	rlf	bin+2,f
	rlf	bin+1,f
	rlf	bin+0,f
	rlf	bcd+4,f		; put it into bcd
	rlf	bcd+3,f
	rlf	bcd+2,f
	rlf	bcd+1,f
	rlf	bcd+0,f
	decfsz	ii,f		; all done?
	goto	b2bcd2		; no, loop
	return			; yes

	end
        

Questions:

• TakeThisOuTbtjlynchKILLspam at SPAMgmail.com asks: " How do you convert an 8 bit unsigned binary number into a BCD number ?
  Brent" James Newton replies: See 8 bit to BCD (half packed) and then just seperate out the nibbles by making a copy, then anding one with 0F and the other with F0 followed by a nibble swap.+

  +

• The code does not appear to work, neither does the code from AN526. I am using a PIC18F452 and have converted the following PIC16 commands: rlf -> rlcf ( rotate left through carry ) FSR -> FSR0L ( setting FSR0H to 0x00, because variables reside in lower memory ) INDF -> INDF0 My only assumption is that i must be doing something really stupid. I tried the code for a bin value of 0xffff and got 06 85 53 ? Can anyone shed some light on this ?+

Interested:

Comments:

• 32-bit binary to BCD (b2bcd) is a very neat algorithm - one thing to note is that the input 32-bit binary is assumed to be big-endian, i.e. most significant byte comes first.+
• TakeThisOuTwalter.pielKILLspam at SPAMsasol.com

  Note that this code as written only works with default radix of 10, otherwise the constants (123456789 and 32), get interpreted as hex numbers. Put a period in front (.123456789 and .32) to make it more idiot proof

  +

Code:

• BIN2DEC	
  	MOVLW	0XFF
  	MOVWF	BIN1
  	MOVLW	0X04
  	MOVWF	BIN2
  	MOVLW	0XFF
  	MOVWF	BIN3
  	MOVLW	0X04
  	MOVWF	BIN4
  	MOVLW	.32		; 32-BITS
  	MOVWF	II		; MAKE CYCLE COUNTER
  	CLRF	DIGIT1
  	CLRF	DIGIT2
  	CLRF	DIGIT3
  	CLRF	DIGIT4
  	CLRF	DIGIT5
  	
  B2BCD2	LFSR	FSR0,DIGIT1		; MAKE POINTER
  	MOVLW	.5
  	MOVWF	CNT
  
  ; MIKE'S ROUTINE:
  
  B2BCD3	MOVLW	0X33		
  	ADDWF	INDF0,F		; ADD TO BOTH NYBBLES
  	BTFSC	INDF0,3		; TEST IF LOW RESULT > 7
  	ANDLW	0XF0		; LOW RESULT >7 SO TAKE THE 3 OUT
  	BTFSC	INDF0,7		; TEST IF HIGH RESULT > 7
  	ANDLW	0X0F		; HIGH RESULT > 7 SO OK
  	SUBWF	INDF0,F		; ANY RESULTS <= 7, SUBTRACT BACK
  	INCF	FSR0L,F		; POINT TO NEXT
  	DECFSZ	CNT
  	GOTO	B2BCD3
  	
  	RLCF	BIN4,F		; GET ANOTHER BIT
  	RLCF	BIN3,F
  	RLCF	BIN2,F
  	RLCF	BIN1,F
  	RLCF	DIGIT5,F		; PUT IT INTO BCD
  	RLCF	DIGIT4,F
  	RLCF	DIGIT3,F
  	RLCF	DIGIT2,F
  	RLCF	DIGIT1,F
  	DECFSZ	II,F		; ALL DONE?
  	GOTO	B2BCD2		; NO, LOOP
  	RETLW	.0		; yes
  +
  

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