Post by thread:3 Byte Binary to BCD Conversion

On Sun, 23 Aug 1998 14:36:59 -0700 Ron Stone <spam_OUTastoneTakeThisOuTEROLS.COM> writes: >Hi -- Does anyone have an assembler routine or algorithm for this? I >have one based on subtraction of 10X >constants but I was wondering if there is something more efficient? >Microchip has one in their application >notes but it's only good for 2 bytes. The "BCD multiply by 2" routine presented in the application note is good for large numbers. It is easy to expand it to more bytes. Increase the loop count to 24 (24 bits in 3 bytes), add another rotate so all 3 bytes of the binary number are rotated, and add more BCD digits and calls to adjbcd. Another method is to repeatedly divide by 10, using the remainders as BCD digits. It takes longer but may save on code space if you already have a suitable divison routine in the program for some other purpose. _____________________________________________________________________ You don't need to buy Internet access to use free Internet e-mail. Get completely free e-mail from Juno at http://www.juno.com Or call Juno at (800) 654-JUNO [654-5866]

Ron Stone wrote: Hi -- Does anyone have an assembler routine or algorithm for this Here is a four byte version. If this doesn't fit your need, compare this to the two byte version and you should be able to figure out how to make the modifications for three bytes. ;Binary - BCD 32 bits (Full Range) ;Input in buff_4|buff_3|buff_2|buff_1, ; ;Converts to packed bcd in temp_a, temp_b, temp_c, temp_d and temp_e ;with the MSD temp_a. ;Handles full range: ff ff ff ff -> 4,294,967,295 ;Also uses temp_f and count. ;2940 cycles including call and return. ;Bob Fehrenbach bin2bcd: bcf STATUS, C movlw 32 movwf count clrf temp_a clrf temp_b clrf temp_c clrf temp_d clrf temp_e bin2bcd_loop: rlf buff_1, f rlf buff_2, f rlf buff_3, f rlf buff_4, f rlf temp_e, f rlf temp_d, f rlf temp_c, f rlf temp_b, f rlf temp_a, f decfsz count, f goto adj_dec goto done adj_dec: movlw temp_e movwf FSR call adj_bcd movlw temp_d movwf FSR call adj_bcd movlw temp_c movwf FSR call adj_bcd movlw temp_b movwf FSR call adj_bcd movlw temp_a movwf FSR call adj_bcd goto bin2bcd_loop adj_bcd: movlw h'3' ;alternative adj_bcd from Mike Keitz addwf INDF, w ;Does not use temp_d movwf temp_f ; movlw h'33' ;Possible correction value. btfsc temp_f, 3 ; addwf INDF,f ;Add to low and high nybbles movwf INDF ; btfsc INDF,3 movlw h'30' ; andlw h'f0' ;Low result >7 . OK (take the 3 out) addwf INDF, w ; btfsc INDF,7 movwf temp_f ; andlw h'0f' ;Hi > 7 OK. btfsc temp_f, 7 ; subwf INDF,f ;any results <=7, subtract back. movwf INDF return done: return

Ron Stone wrote: Hi-Does anyone have an assembler routine or algorithm for this Here is a four byte version. If this doesn't fit your need, compare this to the two byte version and you should be able to figure out how to make the modifications for three bytes. ;Binary - BCD 32 bits (Full Range) ;Input in buff_4|buff_3|buff_2|buff_1, ; ;Converts to packed bcd in temp_a, temp_b, temp_c, temp_d and temp_e ;with the MSD temp_a. ;Handles full range: ff ff ff ff -> 4,294,967,295 ;Also uses temp_f and count. ;2940 cycles including call and return. ;Bob Fehrenbach bin2bcd: bcf STATUS, C movlw 32 movwf count clrf temp_a clrf temp_b clrf temp_c clrf temp_d clrf temp_e bin2bcd_loop: rlf buff_1, f rlf buff_2, f rlf buff_3, f rlf buff_4, f rlf temp_e, f rlf temp_d, f rlf temp_c, f rlf temp_b, f rlf temp_a, f decfsz count, f goto adj_dec goto done adj_dec: movlw temp_e movwf FSR call adj_bcd movlw temp_d movwf FSR call adj_bcd movlw temp_c movwf FSR call adj_bcd movlw temp_b movwf FSR call adj_bcd movlw temp_a movwf FSR call adj_bcd goto bin2bcd_loop adj_bcd: movlw h'3' ;alternative adj_bcd from Mike Keitz addwf INDF, w ;Does not use temp_d movwf temp_f ; movlw h'33' ;Possible correction value. btfsc temp_f, 3 ; addwf INDF,f ;Add to low and high nybbles movwf INDF ; btfsc INDF,3 movlw h'30' ; andlw h'f0' ;Low result >7 . OK (take the 3 out) addwf INDF, w ; btfsc INDF,7 movwf temp_f ; andlw h'0f' ;Hi > 7 OK. btfsc temp_f, 7 ; subwf INDF,f ;any results <=7, subtract back. movwf INDF return done: return

Bob Fehrenbach wrote: ;......... adj_dec: movlw temp_e call adj_bcd movlw temp_d call adj_bcd movlw temp_c call adj_bcd movlw temp_b call adj_bcd movlw temp_a call adj_bcd ;......... adj_bcd: movwf FSR ; movlw h'03' ;alternative adj_bcd from Mike Keitz ; addwf INDF, w ;Does not use temp_d ; movwf temp_f ; movlw h'33' ;Possible correction value. ; btfsc temp_f, 3 ; addwf INDF,f ;Add to low and high nybbles ; movwf INDF ; btfsc INDF,3 movlw h'03' addwf INDF, f btfss INDF, 3 subwf INDF, f Is it not the same ? ; movlw h'30' ; andlw h'f0' ;Low result >7 . OK (take the 3 out) ; addwf INDF, w ; btfsc INDF,7 ; movwf temp_f ; andlw h'0f' ;Hi > 7 OK. ; btfsc temp_f, 7 ; subwf INDF,f ;any results <=7, subtract ; movwf INDF ; ; return The same suggestion. This will save 5*2*32. = 320 clocks and 4+2=6 words of program memory. And one data cell too ;-) WBR Dmitry.

> Ron Stone wrote: > > Hi -- Does anyone have an assembler routine or algorithm for this Another Post replied that you could repeatedly divide by 10 (actually POWERS of 10) and do a BIN to BCD conversion, if you needed to have a divide routine around anyway. I do, so I started working on this. I need a 2 byte (actually, the data in this number will be less than 999 decimal) number converted to 3 byte BCD, not packed BCD. Later, I will need to scale this number (probably by scaling the 2 byte BIN, it's easier) because it represents a Fahrenheit temperature and I need to read out in Centigrade sometimes. So I'll probably need a divide routine. I started messing around with Divide routines today. There is a DBL_DIVS routine in Microchip AN526. It has a note in the header that the Numerator must be greater than the Denominator. I checked this out, and it actually seems to give errors when the Numerator is EQUAL to the denominator as well (I didn't check all cases...) Yuck: 'd103 divided by 'd100 gives Zero remainder 3 not 1 remainder 3 Double Yuck: 'd100 divided by 'd100 gives zero remainder 100 not 1 remainder zero. Anybody got a better routine? I'm looking for a code-efficient (not straight-line) double precision divide routine that will handle all cases. -- Lawrence Lile "The ideal design has zero parts" - (attributed to Harold Hallikainen) Download AutoCad blocks for electrical drafting at: http://home1.gte.net/llile/index.htm

On Mon, 24 Aug 1998 20:42:40 +0400 Dmitry Kiryashov <.....zewsKILLspam@spam@AHA.RU> writes: >Bob Fehrenbach wrote: > >; movlw h'03' ;alternative adj_bcd from Mike Keitz Thanks for the credit, but the code has gotten so unformatted that I can't follow it too well. Let me summarize, since this particular routine comes up every few months. We are looking at the 'adjbcd' process, which is part of the "BCD multiply by 2" binary to BCD routine. What it does is prepare a packed BCD number so that shifting it left will multiply it by 2. This should be done by adding 3 to each digit unless the digit is less than 5 (in which case adding 3 will leave the result less than 8): Digit Adjusted digit Shifted adjusted digit 0 0 0 1 1 2 2 2 4 3 3 6 4 4 8 5 8 0 carry 1 6 9 2 carry 1 7 A 4 carry 1 8 B 6 carry 1 9 C 8 carry 1 Original Microchip (not optimized): movlw h'03' ;Adjust the low digit by adding 3 addwf INDF,w movwf tmp_f ;Store result here for bit test. btfsc tmp_f,3 ;If result < 8, don't change. movwf INDF ;Store the optimized result. movlw h'30' ;Same process on high digit. addwf INDF,w movwf tmp_f btfsc tmp_f,7 movwf INDF Dimitry suggests: movlw h'03' ;Adjust the low digit. addwf INDF,f btfss INDF,3 ;If result >= 8, keep. subwf INDF,f ;Otherwise, de-adjust back. movlw h'30' ;Same for high digit. addwf INDF,f btfss INDF,7 subwf INDF,f My optimized routine: movlw h'33' addwf INDF,f ;Adjust both digits. btfsc INDF,3 ;Low digit <8, adjust back. andlw h'f0' ;Keep the low digit (could use movlw h'30') btfsc INDF,7 ;Hi digit <8, adjust back. andlw h'0f' ;Keep the high digit (by subtracting 0). subwf INDF,f ;Adjust digit(s) back if still 3 in W. So, pick your poison. They all work. Most everyone picked up the optimization of putting the store index value to FSR *inside* the adjbcd routine. For only a few (4 or 6) BCD digits, consider in-lining the adjbcd routine and using direct addressing. For a large number of BCD digits, write a loop to increment FSR and do adjbcd repeatedly. If someone can come up with a way to combine the shifting and adjusting (keeping a proper carry of course), that could further save space on large numbers. _____________________________________________________________________ You don't need to buy Internet access to use free Internet e-mail. Get completely free e-mail from Juno at http://www.juno.com Or call Juno at (800) 654-JUNO [654-5866]

On Mon, 24 Aug 1998 12:08:31 +0000 Lawrence Lile <lilelKILLspamtoastmaster.com> writes: >> Ron Stone wrote: >> >> Hi -- Does anyone have an assembler routine or algorithm for this > > >Another Post replied that you could repeatedly divide by 10 (actually >POWERS of 10) and do a BIN to BCD conversion, if you needed to have a >divide routine around anyway. I do, so I started working on this. No, you divide by 10 every time. The remainders are the BCD digits, but you get them in reverse order. For example, consider 936. 936 div 10 = 93 remainder 6 93 div 10 = 9 remainder 3 9 div 10 = 0 remainder 9 If you want a variable length result, you can stop when the divide result is zero. Or do the process a fixed number of times, and the result will have leading zero digits since 0 div 10 = 0 remainder 0. Put the remainders in order and you have 9 3 6. If you're displaying on an HD44780 LCD screen, you can set it to move the cursor from right to left, and write the digits as they are computed. Otherwise you'd probably need to save them in RAM for output in the usual left to right order. Since the division is always be 10 a simple long division routine can be implemented with just a single byte subtract and test. _____________________________________________________________________ You don't need to buy Internet access to use free Internet e-mail. Get completely free e-mail from Juno at http://www.juno.com Or call Juno at (800) 654-JUNO [654-5866]

Hello Mike Keitz. You gave me another idea how to optimize code. Just idea not been tested. ; buff_1234 values will be corrupted. ; 9 + (12 + 6 + 11 + 11)*31 + 9 + 4 = 1262 clc ; 48 words of program memory bin2bcd: clrf count bsf count,5 ;32. clrf temp_a clrf temp_b clrf temp_c clrf temp_d clrf temp_e goto bin2bcd_loop adj_dec: movlw 0x33 addwf temp_e,f addwf temp_d,f addwf temp_c,f addwf temp_b,f addwf temp_a,f movlw 0x03 btfss temp_e,3 subwf temp_e,f btfss temp_d,3 subwf temp_d,f btfss temp_c,3 subwf temp_c,f btfss temp_b,3 subwf temp_b,f btfss temp_a,3 subwf temp_a,f movlw 0x30 btfss temp_e,7 subwf temp_e,f btfss temp_d,7 subwf temp_d,f btfss temp_c,7 subwf temp_c,f btfss temp_b,7 subwf temp_b,f btfss temp_a,7 subwf temp_a,f bin2bcd_loop: rlf buff_1,f rlf buff_2,f rlf buff_3,f rlf buff_4,f rlf temp_e,f rlf temp_d,f rlf temp_c,f rlf temp_b,f rlf temp_a,f decfsz count, f goto adj_dec return ; 3 bytes variant will require: ; 8 + (10 + 5 + 9 + 9)*23 + 7 + 4 = 778 clc ; 40 words of program memory WBR Dmitry.

On Tue, 25 Aug 1998 11:23:07 +0400 Dmitry Kiryashov <.....zewsKILLspam.....AHA.RU> writes: >You gave me another idea how to optimize code. >Just idea not been tested. It looks like it will work. Probably about the best it can be done in terms of speed (other than unrolling the loop entirely). >; buff_1234 values will be corrupted. If you'd like the binary value to remain the same after conversion, make it shift in a circle like this: rlf buff_4,w ;<-- Add this instr, picks up high bit. rlf buff_1,f rlf buff_2,f rlf buff_3,f rlf buff_4,f rlf temp_e,f etc. After 32 loops, the value in buff will have shifted all the way around, so it would be the same as at the start. The cost is one more instruction which is executed 32 times. > clrf count > bsf count,5 ;32. Cute, but movlw/movwf is the same performance and a lot easier to understand. _____________________________________________________________________ You don't need to buy Internet access to use free Internet e-mail. Get completely free e-mail from Juno at http://www.juno.com Or call Juno at (800) 654-JUNO [654-5866]