BCD question

[keymuu]

I'm not sure how you're testing your Divisor array, but it should not be filled with 3's.

What you should actually be seeing is something like this;

b = 0 : Divisor 0 = 2
b = 1 : Divisor 1 = 11
b = 2 : Divisor 2 = 4
b = 3 : Divisor 3 = 5
b = 4 : Divisor 4 = 3

Divisor[b] should always contain the low byte of the result H/w anded with $0F.

If you just want to extract individual digits from H, then let PBP do it for you with the DIG
operator.

  FOR b = 0 to 4
     divisor [b] = H DIG b
  NEXT b

This divides by 10 returning the remainder.

Example: If H = 37234 then

b = H DIG 0 = 37234/10 = 3723.4 <-- Now b = 4

b = H DIG 1 = 37234/10 = 3723.4. 3723/10 = 372.3 <-- Now b = 3

b = H DIG 2 = 37234/10 = 3723.4. 3723/10 = 372.3. 372/10 = 37.2 <-- Now b = 2

Etc,,

Your entire program gets reduced to this;

Divisor VAR byte[5]
H VAR WORD
b VAR BYTE

Main:
 H=37234 

 'fill divisor with digits 0 to 4 from H
   FOR b = 0 to 4
     divisor [b] = H DIG b
   NEXT b
   GOTO Main

  END

Thank you Bruce

Of course, I forgot the DIG and tried to do the same thing the "basic way", as a algorithm would tell you, so still wondering way it did not work?
Consider that there would not be a DIG statement then the answer wouldn't be as easy as with DIG And then the original idea modified should work, shouldn't it?

Nevertheless, thank you again Bruce for being alert, you have a open mind...

I already tested it with DIG and, of course, it works like ... well... a great thing...