10 bit Binary to BCD Conversion
Posted on the Chat Zone by Ed Grens on September 22, 2004 at 03 08 38
On 18 September Peter Hemsley posted a 12b binary to BCD conversion subroutine
in a response to a post by me. His algorithm is excellent for 12b conversion,
but not efficient for the 10b often encountered with A/D inputs. Over the 10b
data range, 0x000 to 0x3FF, this routine requires on the average of 131
processor cycles. Actually this is more than it requires on the average over the
entire 12b range (118 cycles).
A simple divide by power of ten algorithm, listed below, requires on the average
88 cycles over the 10b range. This is possible because, for 10b, one can use
single byte subtractions for divides. At 12b (probably) and certainly at 16b
this approach cannot compete with the procedure Peter has presented. But, if 10b
is what you need, it can be advantageous.
Ed
;======= Bin2Dec10G.asm ============= Sep 04 =========
; Test of 10b binary to 4 digit conversion as
; based on division by powers of 10.
;
; E. A., Grens
;
; For most PICs - here for 16F628 with 4 MHz
; internal oscillator.
;
; Average processor cycles for execution of
; subroutine over the input range 0x000 to 0x3FF
; is 88. No variables required except the
; input blh, blo and the out BCD digits.
; Subroutine length 38.
;======================================================
list p=16f628
radix dec
#include
__config 0x3D30
;-------Set Up RAM Variables---------------------------
blo equ 0x20 ;lsbyte of 10b binary
bhi equ 0x21 ;msb
d4 equ 0x23 ;msdigit
d3 equ 0x24
d2 equ 0x25
d1 equ 0x26 ;lsd
;-------Program Code--------------------------------
org 0x00 ;Effective Reset Vector
;
nop
goto Start ;Jump by interrupt
;
Start
movlw 0x07
movwf CMCON ;Digital I/O
movlw 0x03 ;Test input
movwf bhi
movlw 0xFF
movwf blo
movlw 0x04
call Bin2decg
Hold
goto Hold ;Finished
Bin2decg ;10b binaey to BCD
clrf d1
clrf d2
clrf d3
clrf d4
clrc
rrf bhi,f ;N = 4*Q + R
rrf blo,f ;blo = Q
rrf d1,f ;d1 = R as temp, R25
goto B2d2 ;repeat
B2d3
addwf blo,f ;get remainder
; after /100, C = 0
rlf d1,f ;*4 and add R back in
rlf blo,f
rlf d1,f
rlf blo,f ;4*blo + R = N -
; 1000*d4 - 100*d3
movlw 10
B2d4
subwf blo,f ;blo = N - 1000*d4 -
; 100*d3 - 10*d2
skpc
goto B2d5 ;blo10
goto B2d4
B2d5
addwf blo,f ;put last 10 back, C=0
movf blo,w
movwf d1
return
end
**********************
12-bit code
Posted by Peter Hemsley on September 18, 2004 at 17 40 56
Message ; Bonus 12-Bit version
; Leaner and Meaner, 36 instructions
; Ideal for displaying A/D values
Bin2DecFast
movf NUMHI,w
iorlw 0xF0 ;w=H2-16
movwf D1 ;D1=H2-16
addwf D1,f ;D1=H2*2-32
addwf D1,f ;D1=H2*3-48
movwf D2 ;D2=H2-16
addlw -D'5' ;w=H2-21
addwf D2,f ;D2=H2*2-37 Done!
addlw D'41' ;w=H2+20
movwf D0 ;D0=H2+20
swapf NUMLO,w
iorlw 0xF0 ;w=H1-16
addwf D1,f ;D1=H2*3+H1-64
addwf D0,f ;D0=H2+H1+4, C=1
rlf D0,f ;D0=(H2+H1)*2+9, C=0
comf D0,f ;D0=-(H2+H1)*2-10
rlf D0,f ;D0=-(H2+H1)*4-20
movf NUMLO,w
andlw 0x0F ;w=H0
addwf D0,f ;D0=H0-(H2+H1)*4-20 Done!
rlf D1,f ;C=0, D1=H2*6+H1*2-128 Done!
movlw D'5'
movwf D3
movlw D'10'
mod0
addwf D0,f ;D(X)=D(X)mod10
decf D1,f ;D(X+1)=D(X+1)+D(X)div10
skpc
goto mod0
mod1
addwf D1,f
decf D2,f
skpc
goto mod1
mod2
addwf D2,f
decf D3,f
skpc
goto mod2
return