b1 = 1 : b2 = 2 : b3 = 3 : b4 = 4 : b5 = 5 : b6 = 6 : b7 = 7 :b8 = 8 : b9 = 9
symbol Hexptr = b10
symbol Sumptr = b11
symbol Incptr = b12
symbol Decval = b13
symbol IncLW = 32 ; Increasing decimal value (in binary format)
symbol SumLW = 40 ; Present decimal value (in binary format)
symbol HexLW = 48 ; Accumulating decimal value (in binary format)
poke IncLW,1 ; Initialise Increment (Higher word must be clear)
poke SumLW,1 ; Initialise Sum (Higher word and Hex words must be clear)
for bptr = 1 to 9
for Decval = 1 to 9
if @bptr = Decval then ; Add decimal value into binary field
Sumptr = HexLW ; Running total
Incptr = SumLW ; value to add
call sum ; Result returned at Sumptr
endif
Sumptr = SumLW
Incptr = IncLW
call sum ; Next decimal value (in binary format),ie: 1,2..9,10,20..90,100,200..
next
sumptr = sumlw
incptr = inclw
peek Sumptr,b0 : poke Incptr,b0 : inc Sumptr : inc Incptr ; Copy sum to next increment
peek Sumptr,b0 : poke Incptr,b0 : inc Sumptr : inc Incptr
peek Sumptr,b0 : poke Incptr,b0 : inc Sumptr : inc Incptr
peek Sumptr,b0 : poke Incptr,b0
next
sertxd("$")
bptr = HexLW + 3
do
b0 = @bptr / 16
lookup b0,("0123456789ABCDEF"),b0
sertxd(b0)
b0 = @bptrdec // 16
lookup b0,("0123456789ABCDEF"),b0
sertxd(b0)
loop until bptr < HexLW
sertxd(cr,lf)
;; Result Should be = 3ADE68B1
end
sum:
peek Sumptr,b14,b15,b16,b17 ; Read w8:w7 (Hi:Lo)
peek Incptr,b18,b19,b20,b21 ; Read w10:w9
w7 = w7 + w9 ; Add Low words
w8 = w8 + w10 ; Add High words
if w7 < w9 then ; Carry to high word
inc w8
endif
poke Sumptr,b14,b15,b16,b17 ; Store sum
returnSomewhat belatedly, I rose to AllyCat’s implied challenge to write faster, more efficient code to achieve this conversion. My program uses the method outlined by AllyCat in #2 (b9*10+b8*10+b7*10...+b1). It has sub-routines for both long word addition and multiplication and only requires 3 bytes of memory (b0 &TempW) in addition to the initial digits (b1-b9), and the long word answer(Word_Hi:Word_Lo) which needs to be initially zero.
#picaxe 08m2
#no_data
symbol Word_Lo=w5
symbol Word_Hi=w6
symbol TempW=w7 'Temporary working word
symbol TempL=b14 'Low byte of working word
symbol TempH=b15 'High byte of working word
symbol PLSB=10 'Pointer to LSB of long word
symbol PMSB=PLSB+3 'Pointer to MSB of long word
b1 = 1 : b2 = 2 : b3 = 3 : b4 = 4 : b5 = 5 : b6 = 6 : b7 = 7 :b8 = 8 : b9 = 9
'b1 = 7 : b2 = 3 : b3 = 5 : b4 = 5 : b5 = 6 : b6 = 0 : b7 = 0 :b8 = 0 : b9 = 0
poke PLSB,b9 'Add most significant digit to long word (assumes long word is zero initially)
for b0=8 to 1 step -1 'Consecutively multiply by 10 and add to long word for each of the remaining digits
call LongWordMult10
bptr=b0
TempH=@bptr
call LongWordPlusByte
next b0
bptr=PMSB 'Print out the result
sertxd("$")
do
b0=@bptr/16
call PrtNib
b0=@bptrdec and $F
call PrtNib
loop until bptr<PLSB
;; Result Should be = 3ADE68B1
end
'-----------------------------------------------------------------
LongWordMult10:
'Multiplies the long word (Word_Hi:Word_Lo)*10
bptr=PLSB
TempH=0
do
TempW=@bptr*10+TempH
@bptrinc=TempL
loop until bptr>PMSB
return
'-----------------------------------------------------------------
LongWordPlusByte:
'Adds TempH to the long word (Word_Hi:Word_Lo)
bptr=PLSB
do
TempW=@bptr+TempH
@bptrinc=TempL
loop while TempH>0
return
'-----------------------------------------------------------------
PrtNib:
'Print the Hex character contained in the low nibble of b0
if b0<10 then
b0=b0+"0"
else
b0=b0+"A"-10
endif
sertxd (b0)
returnThe largest decimal number that can be converted to a 32 bit binary number is 4,294,967,295 so it isn’t possible to convert all 10 digit decimal numbers. If the first digit will never exceed 3 then there won’t be a problem except that space will need to be made to store this extra digit - in b10 presumably which will mean that the location of the long word will need to be moved up a bit.
If you don’t want to get a hex print out of the long word (HiWord & LoWord), simply delete lines 29-37 (commencing with bptr=PMSB....) - the binary number is stored in bytes b10,b11,b12,b13 with b10 containing the LSB.