From: John Payson via Scott Dattalo
see
http://www.dattalo.com/technical/software/pic/bcd.txt
for notes on how this works. Plan on a headache. <GRIN>
[ed: quick guess at speed is that about 200 instructions will be executed
and 50 bytes + 7 registers used]
;Takes hex number in NumH:NumL Returns decimal in ;TenK:Thou:Hund:Tens:Ones ;written by John Payson ;input ;=A_{3}*16^{3} + A_{2}*16^{2} + A_{1}*16^{1} + A_{0}*16^{0} ;=A_{3}*4096 + A_{2}*256 + A_{1}*16 + A_{0} NumH EQU AD3M ;A3*16+A2 NumL EQU AD3L ;A1*16+A0 ;share variables ;=B_{4}*10^{4} + B_{3}*10^{3} + B_{2}*10^{2} + B_{1}*10^{1} + B_{0}*10^{0} ;=B_{4}*10000 + B_{3}*1000 + B_{2}*100 + B_{1}*10 + B_{0} TenK EQU LOOPER ;B_{4} Thou EQU D2 ;B_{3} Hund EQU D1 ;B_{2} Tens EQU R2 ;B_{1} Ones EQU R1 ;B_{0} swapf NumH,w ;w = A_{2}*16+A_{3} andlw 0x0F ;w = A_{3} *** PERSONALLY, I'D REPLACE THESE 2 addlw 0xF0 ;w = A_{3}16 *** LINES WITH "IORLW b'11110000B' " AW movwf Thou ;B_{3} = A_{3}16 addwf Thou,f ;B_{3} = 2*(A_{3}16) = 2A_{3}  32 addlw .226 ;w = A_{3}16  30 = A_{3}46 movwf Hund ;B_{2} = A_{3}46 addlw .50 ;w = A_{3}46 + 50 = A_{3}+4 movwf Ones ;B_{0} = A_{3}+4 movf NumH,w ;w = A_{3}*16+A_{2} andlw 0x0F ;w = A_{2} addwf Hund,f ;B_{2} = A_{3}46 + A_{2} = A_{3}+A_{2}46 addwf Hund,f ;B_{2} = A_{3}+A_{2}46 + A_{2} = A_{3}+2A_{2}46 addwf Ones,f ;B_{0} = A_{3}+4 + A_{2} = A_{3}+A_{2}+4 addlw .233 ;w = A_{2}  23 movwf Tens ;B_{1} = A_{2}23 addwf Tens,f ;B_{1} = 2*(A_{2}23) addwf Tens,f ;B_{1} = 3*(A_{2}23) = 3A_{2}69 (Doh! thanks NG) swapf NumL,w ;w = A_{0}*16+A_{1} andlw 0x0F ;w = A_{1} addwf Tens,f ;B_{1} = 3A_{2}69 + A_{1} = 3A_{2}+A_{1}69 range 69...9 addwf Ones,f ;B_{0} = A_{3}+A_{2}+4 + A_{1} = A_{3}+A_{2}+A_{1}+4 and Carry = 0 (thanks NG) rlf Tens,f ;B_{1} = 2*(3A_{2}+A_{1}69) + C = 6A_{2}+2A_{1}138 and Carry is now 1 as tens register had to be negitive rlf Ones,f ;B_{0} = 2*(A_{3}+A_{2}+A_{1}+4) + C = 2A_{3}+2A_{2}+2A_{1}+9 (+9 not +8 due to the carry from prev line, Thanks NG) comf Ones,f ;B_{0} = ~(2A_{3}+2A_{2}+2A_{1}+9) = 2A_{3}2A_{2}2A_{1}10 (ones complement plus 1 is twos complement. Thanks SD) ;;Nikolai Golovchenko [golovchenko at MAIL.RU] says: comf can be regarded like: ;; comf Ones, f ;; incf Ones, f ;; decf Ones, f ;;First two instructions make up negation. So, ;;Ones = 1 * Ones  1 ;; =  2 * (A3 + A2 + A1)  9  1 ;; =  2 * (A3 + A2 + A1)  10 rlf Ones,f ;B_{0} = 2*(2A_{3}2A_{2}2A_{1}10) = 4A_{3}4A_{2}4A_{1}20 movf NumL,w ;w = A_{1}*16+A_{0} andlw 0x0F ;w = A_{0} addwf Ones,f ;B_{0} = 4A_{3}4A_{2}4A_{1}20 + A_{0} = A_{0}4(A_{3}+A_{2}+A_{1})20 range 215...5 Carry=0 rlf Thou,f ;B_{3} = 2*(2A_{3}  32) = 4A_{3}  64 movlw 0x07 ;w = 7 movwf TenK ;B_{4} = 7 ;B_{0} = A_{0}4(A_{3}+A_{2}+A_{1})20 ;5...200 ;B_{1} = 6A_{2}+2A_{1}138 ;18...138 ;B_{2} = A_{3}+2A_{2}46 ;1...46 ;B_{3} = 4A_{3}64 ;4...64 ;B_{4} = 7 ;7 ; At this point, the original number is ; equal to TenK*10000+Thou*1000+Hund*100+Tens*10+Ones ; if those entities are regarded as two's compliment ; binary. To be precise, all of them are negative ; except TenK. Now the number needs to be normal ; ized, but this can all be done with simple byte ; arithmetic. movlw .10 ;w = 10 Lb1: ;do addwf Ones,f ; B_{0} += 10 decf Tens,f ; B_{1} = 1 btfss 3,0 ;skip no carry goto Lb1 ; while B_{0} < 0 ;jmp carry Lb2: ;do addwf Tens,f ; B_{1} += 10 decf Hund,f ; B_{2} = 1 btfss 3,0 goto Lb2 ; while B_{1} < 0 Lb3: ;do addwf Hund,f ; B_{2} += 10 decf Thou,f ; B_{3} = 1 btfss 3,0 goto Lb3 ; while B_{2} < 0 Lb4: ;do addwf Thou,f ; B_{3} += 10 decf TenK,f ; B_{4} = 1 btfss 3,0 goto Lb4 ; while B_{3} < 0 retlw 0
Archive:
Comments:
btfss 3,0
by btfss STATUS, C, ACCESS
addwf Ones,f ; B0 += 10
decf Tens,f ; B1 = 1
btfss 3,0
decf Tens,f ; B1 = 1
addwf Ones,f ; B0 += 10
btfss STATUS, C, ACCESS
See also:
If you have a 4 digit hexadecimal number, it may be written as N = a_3*16^3 + a_2*16^2 + a_1*16 + a_0 where a_i, i=0,1,2,3 are the four hexadecimal digits. If you wish to convert this to decimal, then you need to express N as N = b_4*10^4 + b_3*10^3 + b_2*10^2 + b_1*10 + b_0 Where b_j, j=0,1,2,3,4 are the five decimal digits. The reason there are 5 digits in the decimal representation is because the maximum four digit hexadecimal number (0xffff) requires 5 decimal digits (65535). Now the goal is to find a set of equations that allow the b_j's to be expressed in terms of the a_i's. There are infinitely many ways to do this. Here are two of probably the simplest expressions. First, expand the 16^i's and then collect all coefficients of the 10^j's: N = a_3*4096 + a_2*256 + a_1*16 + a_0 = a_3*4*10^3 + a_2*2*10^2 + (a_3*9 + a_2*5 + a_1)*10 + 6*(a_3 + a_2 + a_1) + a_0 This gives us five equations: b_0 = 6*(a_3 + a_2 + a_1) + a_0 b_1 = a_3*9 + a_2*5 + a_1 b_2 = 2*a_2 b_3 = 4*a_3 b_4 = 0 Which as John says, must be "normalized". Normalization in this context means we need to reduce the b_j's such that 0 <= b_j <= 9 In other words we need to find: c_0 = b_0 mod 10 b_1 = (b_1 + (b_0  c_0)/10) c_1 = b_1 mod 10 b_2 = (b_2 + (b_1  c_1)/10) c_2 = b_2 mod 10 b_3 = (b_3 + (b_2  c_2)/10) c_3 = b_3 mod 10 c_4 = (b_4 + (b_3  c_3)/10) mod 10 = (b_2  c_2)/10 Division by 10 can be done quite efficiently (as was shown in another thread several months ago). However, it does require a significant amount of code space compared to say repeated subtractions. Unfortunately, there can be very many subtractions that are required. For example, b_1 could be as large as 15*16, or 240. 10 would have to be subtracted 24 times if you wish to compute 240 mod 10. I presume John realized this inefficiency and thus sought to express N so that repeated subtractions could be used and that the total number of subtractions are minimized. This leads to the next way that N can be expressed: N = a_3*(4100  4) + a_2*(260  4) + a_1*(204) + a_0 = 4*a_3*10^3 + (a_3 + 2*a_2)*10^2 + (6*a_2 + 2*a_1)*10 + a_0  4*(a_3 + a_2 + a_1) This gives five new equations for the b_j's. b_0 = a_0  4*(a_3 + a_2 + a_1) b_1 = 6*a_2 + 2*a_1 b_2 = a_3 + 2*a_2 b_3 = 4*a_3 b_4 = 0 However, these equations are still not conducive to the repeated subtraction algorithm, at least the way John has done it. In other words, it is possible to precondition each of the b_j's so that they are less than zero. Then the repeated subtractions can simultaneously perform the "mod 10" and "/10" operations shown above. Consider the equation b_0 for example, b_0 = a_0  4*(a_3 + a_2 + a_1) Since each a_i must satisfy: 0 <= a_i <= 15, then b_0 ranges: 60 <= b_0 <= 15 We can make b_0 negative by subtracting any number greater than 15. A logical choice is 20. This is because if we subtract 20 from b_0, we can add 2 to b_1 to keep the net result the same. The reason we add "2" can be seen: b_1*10 + b_0 = b_1*10 + b_0 + 20  20 = (b_1 + 2)*10 + b_0  20 Carrying this concept out for the rest of the b_i's we have. b_0 = a_0  4*(a_3 + a_2 + a_1)  20 b_1 = 6*a_2 + 2*a_1 + 2  140 = 6*a_2 + 2*a_1  138 b_2 = a_3 + 2*a_2 + 14  60 = a_3 + 2*a_2  46 b_3 = 4*a_3 + 6  70 = 4*a_3  64 b_4 = 0 + 7 = 7 And if you look at John's code closely, you will see this is how he has expressed the b_j's.+
file: /Techref/microchip/math/radix/b2bu16b5d.htm, 11KB, , updated: 2011/10/16 19:58, local time: 2019/10/13 16:14,

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