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PIC Microcontoller Radix Math Method


24bit to BCD unpacked 8 digits (several versions) in as little as 25 inst or as few as 1242 cycles by Peter Hemsley, Steve Teal, Harry West, and Michael McLoughlin

BETTER BIN-DEC CONVERSION (Peter Hemsley - Sept '00)

This binary to decimal routine is neat and fast and 24-bit but easily modified to 16 or 32-bit. Execution time is constant and so can be used where timing is critical.

I got the idea from the way some processors execute a decimal adjust instruction in hardware, so did a bit of simple arithmetic and some lateral thinking. This version is my generic one, no real need for the subroutines unless they are called from elsewhere. In the 16-bit version I expanded the two inner loops, the resulting code is hardly any bigger, executes faster, uses only one loop counter and does not use the FSR. Great for the smaller PICs. An R Pere replies: hello i am trying to copy and paste and you source code doesnt work for me 24 bits to bcd 8 digits i am very interrenting my email is antoniorubiojose@gmail.com thanks for help me+

+

BINDEC:  	CALL CLRDIG
        	MOVLW 24
        	MOVWF COUNTER1
        	GOTO SHIFT1
ADJBCD:  	MOVLW DIGIT1
        	MOVWF FSR
        	MOVLW 7
        	MOVWF COUNTER2
        	MOVLW 3
ADJLOOP: 	ADDWF INDF,F
        	BTFSS INDF,3
        	SUBWF INDF,F
        	INCF FSR,F
        	DECFSZ COUNTER2,F
        	GOTO ADJLOOP
SHIFT1:  	CALL SLCNT
SLDEC:   	MOVLW DIGIT1
        	MOVWF FSR
        	MOVLW 8
        	MOVWF COUNTER2
SLDLOOP:
        	RLF INDF,F
        	BTFSC INDF,4
        	BSF STATUS,C
        	BCF INDF,4
        	INCF FSR,F
        	DECFSZ COUNTER2,F
        	GOTO SLDLOOP
        	DECFSZ COUNTER1,F
        	GOTO ADJBCD
        	RETURN
SLCNT:   	RLF COUNT0,F
        	RLF COUNT1,F
        	RLF COUNT2,F
        	RETURN
CLRDIG:  	CLRF DIGIT1
        	CLRF DIGIT2
        	CLRF DIGIT3
        	CLRF DIGIT4
        	CLRF DIGIT5
        	CLRF DIGIT6
        	CLRF DIGIT7
        	CLRF DIGIT8
        	RETURN

;..........
+

DECIMAL CHALLENGE (Steve Teal - Nov '00)

Peter Hemsley’s excellent solution to the age-old and often messy problem of converting binary information into human perceivable decimal inspired me to develop my own solution.

A multiply by two with carry is performed on each digit, nested inside a bit counter, each bit determining the state of the carry for the first multiply. This routine takes 1,957 instruction cycles to execute which is a slight improvement on Peter’s, which uses 3,086. Both routines maintain a constant execution time. It will be just as easy to change the number of bits converted.

The commands "movlw -10, addwf INDF, W"’ are for the PICs that lack any subtract instructions, the 16c5x for example, "movlw 10, subwf INDF, W’’ works just the same. +

BIN2DEC		clrf DEC0      	; Clear decimal output buffer
	 	clrf DEC1
		clrf DEC2
	 	clrf DEC3
		clrf DEC4
		clrf DEC5
		clrf DEC6
		clrf DEC7       ; NB. BINn and FSR are trashed after this routine
		movlw 24        ; Initiate bit loop
		movwf BIT_COUNTER
BITLOOP 	rlf BIN0, F	; Every iteration of this loop will copy the next
        	rlf BIN1, F	; bit of the bin value, starting with the MSB,
        	rlf BIN2, F	; to the carry flag
        	movlw DEC0
        	movwf FSR	; Initiate DECn pointer and counter
        	movlw 8
        	movwf DEC_COUNTER	; The following is executed 8 times per bit
DECLOOP 	rlf INDF, F	; Multiply DECn by two with carry, DECn * 2 + C
        	movlw -10	; See note above - test for DECn > 9
        	addwf INDF, W	; W = DECn -10, if W = positive or zero, C = 1
        	btfsc STATUS, C	; DECn has overflowed (>>9) if carry is set
        	movwf INDF	; If carry is set DECn = DECn - 10
        	incf FSR, F	; Carry is CARRIED over to next multiply
        	decfsz DEC-COUNTER, F
        	goto DECLOOP	; Multiply next DECn
        	decfsz BIT-COUNTER, F
        	goto BITLOOP	; Do next bit
        	retlw 0		; Could be RETURN on most PICs

;.............
+

BCD CHALLENGE ACCEPTED (Harry West - Dec '00)

I was very intrigued by the Binary to BCD conversion routine given in Sept '00, as I had always seen this done by some method involving division by ten.

After a lot of thought, I managed to come up with what I think is an improved version. The procedure used to do the conversion is: "Start with a Partial Result (PR) of zero. For each bit in the binary, starting at the left hand end, multiply the PR by two and add the bit."

By doing this arithmetic in decimal, the PR at the end has the converted value which holds the digits as binary coded decimal (BCD) in the lower four bits of each of a succession of bytes, bits 4 to 7 are zeroes (Unpacked BCD). You could also, with a different program, use Packed BCD with two digits in a byte, one in each nibble.

Throughout the process, decimal adjustment (DecAdj) of the PR is necessary to maintain its BCD nature, so that 0 to 9 are unchanged but a result in the range 10 to 15, which is stored as hex 0A to 0F, is converted to 0 to 5 with a 1 carry ready to go in the next BCD, i.e. 0A to 0F become 10 to 15 hex.

The actual process is to add six to the unadjusted result. If this causes a 1 in the fifth bit (bit 4) then the changed pattern is used, otherwise the original unadjusted pattern is retained. For Unpacked BCD the state of bit 4 can be used as the test.

The algorithm in Sept '00 uses "Add 3" before the "times 2" shift. This is best when Packed BCD is converted since for the "top" nibble there is no bit corresponding to bit 4. By "Adding 3" before the shift instead of "Add 6" after, the same effect is obtained using bit 7. However, in this case the carry into the decimal cannot be done until after the shift, hence the two passes through the digits for each bit.

The following is a version using one pass, for Unpacked BCD. I have used the locations BIN0 to BIN2 to hold the three bytes of binary, with the most significant (m.s.) byte in BIN2. The PR goes in the eight bytes DIGIT0 to DIGIT7, with the m.s. digit in DIGIT7. BINCNT and DECCNT hold counts for the two nested loops. +

BINDEC:
	CALL CLRDIG    ; Clear decimal digits
	MOVLW 24       ; Decimal count
	MOVWF BINCNT

BITLP:
	RLF BIN0,F     ; Shift binary left
	RLF BIN1,F     
	RLF BIN2,F
	MOVLW DIGIT0
	MOVWF FSR
	MOVLW 8        ; Count for the decimal digits
	MOVWF DECCNT
	MOVLW 6        ; The Working Register holds 6 throughout. For each bit the inner loop is repeated 8 times, with shift in of the next bit, "times 2" and DecAdj of each digit

ADJLP:
	RLF INDF,F     ; 2*digit, then shift in "next bit’’ for DIGIT0 or else the carry from the previous digit
	ADDWF INDF,F   ; Add 6, clears Cf and gives 1 in bit 4 if the
	BTFSS INDF,4   ; addition is needed; zero if not, when
	SUBWF INDF,F   ; we subtract it again. Sets Cf
	BSF STATUS,C   ; Cf could be 0 or 1, so make it 1 as default
	BTFSS INDF,4   ; Bit 4 is the carry to the next digit
	BCF STATUS,C   ; Reset Cf to zero if bit 4 is clear
	BCF INDF,4     ; For BCD clear bit 4 in case it’s one
	INCF FSR,F     ; Go to next digit, (Cf not affected)
	DECFSZ DECCNT,F ; End of inner loop. check digit count and
	GOTO ADJLP     ; round again if it’s not zero
	DECFSZ BINCNT,F ; End of outer loop, one pass through digits,
	GOTO BITLP     ; check bit count and repeat if necessary.
	RETURN


;..........
+

SHORTER BCD CONVERSION (Michael McLoughlin - April '01)

I have modified Steve Teal’s code, which required 1957 cycles, so that it completes in 1242 cycles!

Steve’s code doubles his Travelling Total (TT), but this only grows slowly and initially only one digit is needed to handle it. Yet the subroutine always doubles the whole of TT, so almost half the RLF multiplications do 2 × 0 + 0 = 0, and are superfluous. By studying the worst case (all 24 bits = 1) it soon appears that we only need to involve a new decimal digit for every three binary digits. The 08 in Steve’s listing could be called cycles, to start at 01 and increment after every three bits. Another counter (octcnt) ensures the repetition of that whole procedure just eight times.

In the following listing the commands to delete are shown "remmed out" with a semicolon, and the new lines are notated as such. +

bin2dec:
    clrf dec0
    clrf dec1
    cIrf dec2
    cIrf dec3
    cIrf dec4
    cIrf dec5
    cirf dec6
    cIrf dec7
;   movlw 18	 ; deleted
    clrf cycles	 ; new
    movlw 08	 ; new
    movwf octcnt ; new

ctloop:	 	 ; new
    incf cycles  ; new
    movlw 03   	 ; new
    movwf bitcnt

bitloop:
    rlf bin0
    rlf bin1
    rlf bin3
    movlw dec0
    movwf FSR
;   movlw 08	 ; deleted
    movfw cycles ; new
    movwf deccnt

decloop:
    rlf INDF
    movlw 0xF6
    addwf INDF,0
    btfsc STATUS,0
    movwf INDF
    incf FSR
    decfsz deccnt
    goto decloop
    decfsz bitcnt
    goto bitloop
    decfsz octcnt ; new
    goto octloop  ; new
    return

;............

;From Peter Hemsley 15July01

;Binary to decimal
;Convert 24 bit binary (count0,1,2) to decimal digits 1 to 8
;Revised version. My thanks to all who contributed to the bin to dec
;discussion in EPE.

	processor 16f84
	include p16f84.inc
	radix dec

;Ram equates
count0	equ	0x0C		;lsb
count1	equ	0x0D
count2	equ	0x0E		;msb
digit1	equ	0x11		;lsd
digit2	equ	0x12
digit3	equ	0x13
digit4	equ	0x14
digit5	equ	0x15
digit6	equ	0x16
digit7	equ	0x17
digit8	equ	0x18		;msd
bitcnt	equ	0x19
digcnt	equ	0x1A


	org	0		;test code for MPLAB simulator
	movlw	0xFF
	movwf	count2
	movlw	0xFF
	movwf	count1
	movlw	0xFF
	movwf	count0
	call	bin2dec
brk1	return


bin2dec	call	clrdig
	movlw	24		;24 bits to do
	movwf	bitcnt

bitlp	rlf	count0		;Shift msb into carry
	rlf	count1
	rlf	count2

	movlw	digit1
	movwf	fsr		;Pointer to digits
	movlw	8		;8 digits to do
	movwf	digcnt
adjlp	rlf	indf		;Shift digit 1 bit left
	movlw	-10
	addwf	indf,w		;Check and adjust for decimal overflow
	skpnc
	movwf	indf

	incf	fsr		;Next digit
	decfsz	digcnt
	goto	adjlp
	decfsz	bitcnt		;Next bit
	goto	bitlp
	return

clrdig:	clrf	digit1
	clrf	digit2
	clrf	digit3
	clrf	digit4
	clrf	digit5
	clrf	digit6
	clrf	digit7
	clrf	digit8
	return

	end
+

See also:

Questions:

Comments:


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