From: John Payson via Scott Dattalo
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 ;=A3*163 + A2*162 + A1*161 + A0*160 ;=A3*4096 + A2*256 + A1*16 + A0 NumH EQU AD3M ;A3*16+A2 NumL EQU AD3L ;A1*16+A0 ;share variables ;=B4*104 + B3*103 + B2*102 + B1*101 + B0*100 ;=B4*10000 + B3*1000 + B2*100 + B1*10 + B0 TenK EQU LOOPER ;B4 Thou EQU D2 ;B3 Hund EQU D1 ;B2 Tens EQU R2 ;B1 Ones EQU R1 ;B0 swapf NumH,w ;w = A2*16+A3 andlw 0x0F ;w = A3 *** PERSONALLY, I'D REPLACE THESE 2 addlw 0xF0 ;w = A3-16 *** LINES WITH "IORLW b'11110000B' " -AW movwf Thou ;B3 = A3-16 addwf Thou,f ;B3 = 2*(A3-16) = 2A3 - 32 addlw .226 ;w = A3-16 - 30 = A3-46 movwf Hund ;B2 = A3-46 addlw .50 ;w = A3-46 + 50 = A3+4 movwf Ones ;B0 = A3+4 movf NumH,w ;w = A3*16+A2 andlw 0x0F ;w = A2 addwf Hund,f ;B2 = A3-46 + A2 = A3+A2-46 addwf Hund,f ;B2 = A3+A2-46 + A2 = A3+2A2-46 addwf Ones,f ;B0 = A3+4 + A2 = A3+A2+4 addlw .233 ;w = A2 - 23 movwf Tens ;B1 = A2-23 addwf Tens,f ;B1 = 2*(A2-23) addwf Tens,f ;B1 = 3*(A2-23) = 3A2-69 (Doh! thanks NG) swapf NumL,w ;w = A0*16+A1 andlw 0x0F ;w = A1 addwf Tens,f ;B1 = 3A2-69 + A1 = 3A2+A1-69 range -69...-9 addwf Ones,f ;B0 = A3+A2+4 + A1 = A3+A2+A1+4 and Carry = 0 (thanks NG) rlf Tens,f ;B1 = 2*(3A2+A1-69) + C = 6A2+2A1-138 and Carry is now 1 as tens register had to be negitive rlf Ones,f ;B0 = 2*(A3+A2+A1+4) + C = 2A3+2A2+2A1+9 (+9 not +8 due to the carry from prev line, Thanks NG) comf Ones,f ;B0 = ~(2A3+2A2+2A1+9) = -2A3-2A2-2A1-10 (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 ;B0 = 2*(-2A3-2A2-2A1-10) = -4A3-4A2-4A1-20 movf NumL,w ;w = A1*16+A0 andlw 0x0F ;w = A0 addwf Ones,f ;B0 = -4A3-4A2-4A1-20 + A0 = A0-4(A3+A2+A1)-20 range -215...-5 Carry=0 rlf Thou,f ;B3 = 2*(2A3 - 32) = 4A3 - 64 movlw 0x07 ;w = 7 movwf TenK ;B4 = 7 ;B0 = A0-4(A3+A2+A1)-20 ;-5...-200 ;B1 = 6A2+2A1-138 ;-18...-138 ;B2 = A3+2A2-46 ;-1...-46 ;B3 = 4A3-64 ;-4...-64 ;B4 = 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 ; B0 += 10 decf Tens,f ; B1 -= 1 btfss 3,0 ;skip no carry goto Lb1 ; while B0 < 0 ;jmp carry Lb2: ;do addwf Tens,f ; B1 += 10 decf Hund,f ; B2 -= 1 btfss 3,0 goto Lb2 ; while B1 < 0 Lb3: ;do addwf Hund,f ; B2 += 10 decf Thou,f ; B3 -= 1 btfss 3,0 goto Lb3 ; while B2 < 0 Lb4: ;do addwf Thou,f ; B3 += 10 decf TenK,f ; B4 -= 1 btfss 3,0 goto Lb4 ; while B3 < 0 retlw 0
btfss STATUS, C, ACCESS
addwf Ones,f ; B0 += 10
decf Tens,f ; B1 -= 1
decf Tens,f ; B1 -= 1
addwf Ones,f ; B0 += 10
btfss STATUS, C, ACCESS
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*(20-4) + 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 pre-condition 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.+
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