BCD question

Hi!

Consider this code:

Code:

---

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

powerTen:            ' for BCD conversion
  select case b
    case 0: w = 1
    case 1: w = 10
    case 2: w = 100
    case 3: w = 1000
    case 4: w = 10000
  end select
return

 H=37234

 'convert to BCD and fill divisor
        FOR b = 0 to 4
            gosub powerTen
            divisor [b] = (H/w) & $0f
        next b
      ENDIF

---

Suspecting that "(H/w) & $0f" does not work as one would think, that is
w=1 -> 37234/1 = 37234 & $F -> 4
w=2 -> 37234/10= 3723.4 & $F = 3723 -> 3
w=3 -> 37234/100=372.34 & $F= 372 -> 2
w=4 -> 37234/1000=37,234 & $F=37 -> 7
w=5 -> 37234/10000=3.7234 & $F=3 -> 3

The array is filled with 3, that is 33333 !:mad:

The goal is to find in that array 37234...

So I fail to understand what is wrong and how to amend? :o

Can anybody explain why that code doesn't work

Thanks :)

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;

Code:

---

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.

Code:

---

  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;

Code:

---

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

---

Quote:

---

Originally Posted by Bruce

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;

Code:

---

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.

Code:

---

  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;

Code:

---

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 :rolleyes: And then the original idea modified should work, shouldn't it?

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

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

You can do the same thing without DIG.

Code:

---

Main:
 H=37234

 'fill divisor with digits 0 to 4 from H
  FOR b = 0 to 4
    divisor[b] = H//10
    H=H/10 
  NEXT b
  GOTO Main

---

As a learning exercise, see if you can figure out why this works the same as DIG - or come
up with another method...;o}

Quote:

---

Originally Posted by Bruce

You can do the same thing without DIG.

Code:

---

Main:
 H=37234

 'fill divisor with digits 0 to 4 from H
  FOR b = 0 to 4
    divisor[b] = H//10
    H=H/10 
  NEXT b
  GOTO Main

---

As a learning exercise, see if you can figure out why this works the same as DIG - or come
up with another method...;o}

---

Ok, Bruce you are also a teacher :)

As earlier said, I did not remember the DIG statement and obviously not modulus divider "//" neither.

That piece of code is clear and surely should work, however, if wouldn't work as expected then again there would be something to marvel over ;) But fortunately that is not the case....

So if we compare the "original" and your excellent DIG interpreter:

Code:

---

a)
    FOR b = 0 to 4
      gosub powerTen
      divisor [b] = (H/w) & $0f
    next b
b)
  FOR b = 0 to 4
    divisor[b] = H//10
    H=H/10 
  NEXT b

---

Still wondering what is wrong in a?
If we convert those code fragments to algorithms then them would be something like this:
a) divide H with power of ten (10^n, when goes n=0..4) to get rid of the decimals, isolate the lsd and store it. Do it five times.
b) isolate the lsd with H mod 10, store it and truncate H with one digit. Do it five times. Simple and clear algorithm....

I just don't grasp what is wrong with a, I do not need it anymore, just wondering :confused:

Quote:

---

Still wondering what is wrong in a?

---

OK let's work through this a bit to see what's wrong with a.

On the 1st pass;

H = 37234
w = 1

37234/1 = 37234. Not much help so far. This is $9172 in HEX.

$9172 has a high byte of %10010001 & a low byte of %01110010.

Since Divisor is a byte array, Divisor[b] will always be loaded with
the low byte of the result (H/w) & $0F.

Now take your low byte %01110010 & AND it with %00001111. The result is
%00000010 or 2d, which, of course, isn't what you're expecting.

Do you see how this is working?

Not yet! OK we'll do one more;

on the 2nd pass;

H = 37234
w = 10

37234/10 = 3723.4. The .4 is lost so you have 3723, which = $E8B.

$E8B has a high byte of %00001110 & a low byte of %10001011.

Now take your low byte %10001011 & AND it with %00001111. The result is
%00001011 or 11d, which again is not what you're expecting.

OK - one more pass;

On the 3rd pass H still = 37234 and w = 100.

37234/100 = 372.34. Toss the .34 and you're left with 372, which = $174.
The high byte is %00000001 the low byte is %01110100.

The low byte %01110100 & %00001111 = %00000100 or 4d. You expected a
2d, but it returned 4d. Now you know why...;o}

That's how a is working. Just follow it along. It's really simple if you just take it
apart step-by-step.