From: Mike Mansheim of Graco, Inc.

/*

- based on the 16 bit to 5 digit conversion written by John Payson & explained by Scott Dattalo - this version adds the extra bit so that the 5 digit display can read to 99,999 - want to express the b_i's in terms of the a_i's for the equations: N = a4*16^4 + a3*16^3 + a2*16^2 + a1*16 + a0 N = b4*10^4 + b3*10^3 + b2*10^2 + b1*10 + b0 the solution coded below: b0 = a0 - (4 * (a4 + a3 + a2 + a1)) - 20 b1 = (4 * a4) + (6 * a2) + (2 * a1) - 198 b2 = (5 * a4) + a3 + (2 * a2) - 140 b3 = (5 * a4) + (4 * a3) - 144 b4 = (6 * a4) + 16 - written in CCS C - since ram was plentiful, I used a 3 byte number - only one bit of the extra byte is actually required. I also used three bytes of scratchpad ram to increase the efficiency of the compiled code. The equations can be coded in C exactly as shown above, but that doesn't take advantage of common terms & progressive multiplies. - even with just the one extra bit, the incoming number can exceed 99,999, so it needs to be checked. This code doesn't show that. - CCS C is limited to 16 bit as the largest integer data type. To use a 24 bit number, I use an array to make sure the bytes are consecutive. This function operates on the array as a global, because I think that is more efficient than passing arrays to the function. - the code could be quite a bit smaller - in John's original, the portion of the code that isolated the nibbles and calculated the digits took 32 instructions. If I code that solution in C, using two scratchpad bytes to improve things, it compiles to 64 instructions. This version compiles to 85 instructions, so I'm sure an asm whiz could show quite an improvement. The C version is still much faster than the the original way I did this, and is adequate for what I need. */ unsigned int TenK, Thou, Hund, Tens, Ones; unsigned int num[3]; void BIN2DEC(void) { // operates on global variables: // set num[0]-[3] (lo-hi) before calling // returns with TenK...Ones set from 0-9 unsigned int a0, a1, a2, a3, a4; unsigned int t1, t2, t3; // scratchpad variables // isolate the 4 bit nibbles // for the purposes of this routine, a4 can be only 0 or 1 a4 = num[2]; a3 = num[1] / 16; a2 = num[1] & 0b00001111; a1 = num[0] / 16; a0 = num[0] & 0b00001111; // calculate the decimal digits (the b_i's are expressed here // as TenK...Ones, as in the original) // all are negative, except TenK t1 = 4 * a4; Ones = a0 - t1 - 20; t2 = 2 * a1; Tens = t1 + t2 - 198; t3 = 2 * a2; Hund = a3 + t3 - 140; t3 = t3 * 2; // now = 4 * a2 t2 = t2 * 2; // now = 4 * a1 Ones = Ones - t3 - t2; t3 = t3 + a2 + a2; // now = 6 * a2 Tens = Tens + t3; // Tens done t1 = t1 + a4; // now = 5 * a4 Hund = Hund + t1; t3 = 4 * a3; Thou = t1 + t3 - 144; // Thou done Ones = Ones - t3; // Ones done TenK = t1 + a4 + 16; // Tenk done // "normalize" the digits - this asm code was // copied directly from the original #asm movlw 0x0A Lb1: addwf Ones,f decf Tens,f btfss 3,0 goto Lb1 Lb2: addwf Tens,f decf Hund,f btfss 3,0 goto Lb2 Lb3: addwf Hund,f decf Thou,f btfss 3,0 goto Lb3 Lb4: addwf Thou,f decf TenK,f btfss 3,0 goto Lb4 #endasm }anhalt at @spam@zoominternet.net

I tweaked Mike Mansheim's algorithm a little and would like to offer this code:+

PICLIST Submission

17BIN2BCD

;February 6, 2018 ;John Anhalt ;41 Tcy to Normalize ;Enhanced mid-range PIC16F1829 ;"#" marks final polynomial for each BCD cblock 0x20 binU binH binL ones tens hund thou tenK endc org 0x0000 nop bra Start Start DoA2 movf binH,w andlw 0x0F ;w=a2 movwf ones ;ones=a2 movwf tens ;tens=a2 lslf WREG ;w=2a2 addlw 0xC7 ;-57 movwf hund ;hund=2a2-57 addlw 0xF9 ;-7 -57= -64 addwf tens,f ;tens=3a2-64 DoA4 movf binU,w addwf ones,f ;ones=a4+a2 addwf hund,f ;hund=a4+2a2-57 movwf thou ;thou=a4 lslf WREG ;w=2a4 movwf tenK ;tenK=2a4 addwf tens,f ;tens=2a4+3a2-64 lslf WREG ;w=4a4 addwf hund,f ;hund=5a4+2a2-57 addwf thou,f ;thou=5a4 addlw 8 ;w=4a4+8 addwf tenK,f ;tenK=6a4+8 # DoA3 swapf binH,w andlw 0x0F ;w=a3 addwf ones,f ;ones=a4+a3+a2 addwf hund,f ;hund=5a4+a3+2a2-57 # lslf WREG ;w=2a3 lslf WREG ;w=4a3 addlw 0xB7 ;-73 addwf thou,f ;thou=5a4+4a3-73 # DoA1 ;same as Payson/Dattalo, binL preserved swapf binL,w andlw 0x0F ;w=a1 addwf tens,f ;tens=2a4+3a2+a1-64 addlw 4 ;see: Payson/Dattalo explanation addwf ones ;ones=a4+a3+a2+a1+4 lslf tens,f ;tens=4a4+6a2+2a1-128,C is set # rlf ones,f ;moves C into ones comf ones,f rlf ones,f ;w/ carry ones is complement 2 negated movf binL,w andlw 0x0F ;w=a0 addwf ones,f ;ones=a0-4(a4+a3+a2+a1)-20 # Normalize movlw 0x0A Lb1 addwf ones,f decf tens,f btfss STATUS,0 bra Lb1 Lb2 addwf tens,f decf hund,f btfss STATUS,0 bra Lb2 Lb3 addwf hund,f decf thou,f btfss STATUS,0 bra Lb3 Lb4 addwf thou,f decf tenK,f btfss STATUS,0 bra Lb4 bra Done DoneExplanation (Based on Mike Mansheim)

Consider a 17-bit binary (N)= a4(2^16) +a3(2^12) + a2(2^8) + a1(2^4) +a0(2^0) where a_i is a 4-bit nibble.

A set of polynomials can be derived where b_i is the BCD digit:

b0(ones) = a0-4(a4+a3+a2+a1)

b1 (tens) = 4a4+6s2+2a1

b2(hund) = 5a4+a3+2a2

b3(thou) = 5a4+4a3

b4(tenK) = 6a4

Now include adjustment to ensure each equation, except tenK, is negative:

b0(ones) = a0-4(a4+a3+a2+a1)-20

b1 (tens) = 4a4+6s2+2a1-128

b2(hund) = 5a4+a3+2a2-57

b3(thou) = 5a4+4a3-73

b4(tenK) = 6a4+8

The above polynomials are slightly different from those derived by Mansheim and result in code that runs a little faster. Since this set is for only 17 bits, the maximum value for a_4 was 1. John Anhalt

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