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I recently became the owner of HP-16c and I want make a program to convert hexadecimal number to decimal by a special rule.
I use the following formulas in Excel:

1. Remove in the hexadecimal number the last two digits from left:
`=LEFT(A2,(LEN(A2)-2))`
2. Converts HEX to DEC:
`=HEX2DEC(B2)`
3. Remove first digits starting with the “1“ for such a rule
```=IF(LEFT(C2,4)="1000",RIGHT(C2,1),
(IF(LEFT(C2,3)="100",RIGHT(C2,2),
(IF(LEFT(C2,2)="10",RIGHT(C2,3),
(IF(LEFT(C2,1)="1",RIGHT(C2,4),
"???")))))))```
For example:
```HEX	DEC
285801	328
271101	1
3e7F03	5999
3f7F03	6255
ae01	174
f02	15
3dA01	???
3f3e02	6190
2b6799	1111```
But I have difficulty trying to program HP-16c begin at first step-how to remove two digit? Shifting Bits, or...?

That part is easy. Simply shift (not rotate) right eight bits. Converting from hex to decimal is also easy.
Stripping off leading ones from a decimal number is somewhat harder. Divide by 1000, see if the quotient is 1, divide by 100, see if the quotient is 1, etc.

Or maybe:

• Subtract 1000
• Leave if it is between 0 and 1000

• Subtract 100
• Leave if it is between 0 and 100

(...)

This is the program that illustrates the algorithm:

```#!/usr/bin/python
def remove(n):
k = 1000
while k > 0:
n -= k
if 0 <= n < k:
break
n += k
k /= 10
return n
for n in (0, 1, 3, 13, 45, 114, 144, 456, 1456, 1018, 3255):
print "%10d -> %10d" % (n, remove(n))

0 ->          0
1 ->          0
3 ->          3
13 ->          3
45 ->         45
114 ->         14
144 ->         44
456 ->        456
1456 ->        456
1018 ->         18
3255 ->       3255
```

This is the program for the HP-16C:
```001 - 43,22, A  LBL A
002 -        1  1
003 -        0  0
004 -        0  0
005 -       10  /
006 -       24  DEC
007 -        1  1
008 -        0  0
009 -        0  0
010 -        0  0
011 -        0  0
012 - 43,22, 0  LBL 0
013 -       30  -
014 -    43  2  x<0
015 -    22  1  GTO 1
016 -    43 36  LSTx
017 -    43  1  x<=y
018 -    22  2  GTO 2
019 -    22  3  GTO 3
020 - 43,22, 1  LBL 1
021 -    43 36  LSTx
022 - 43,22, 2  LBL 2
023 -       40  +
024 -    43 36  LSTx
025 -        1  1
026 -        0  0
027 -       10  /
028 -    43 30  x>0
029 -    22  0  GTO 0
030 - 43,22, 3  LBL 3
031 -       33  Rv
032 -    43 21  RTN
```

My table doesn't agree in all cases:

```INPUT         OUTPUT
285801 h       328 d
271101 h         1 d
3E7F03 h      5999 d
3F7F03 h      6255 d
AE01 h        74 d
F02 h         5 d
3DA01 h       986 d
3F3E02 h      6190 d
2B6799 h      1111 d
```

Nevertheless I hope this is what you were looking for.

Kind regards

Thomas

Edited: 3 June 2013, 2:58 p.m.

Quote:
Stripping off leading ones from a decimal number

As the examples ae01 and f02 indicate that's not what Leonid wants. But then I wonder, what's happening with this formula in Excel with say 100037: is it just 7 as IF(LEFT(C2,4)="1000",RIGHT(C2,1) indicates? But then this isn't the description:

Quote:
Remove first digits starting with the “1“ for such a rule

Or is the length of the decimal numbers somehow limited to 5 digits? Do we just have to remove the first digit if the decimal number exceeds 9999? This program produces the same results as your table with the exception of 3DA01 h for which it returns 986 d:
```001 - 43,22, A  LBL A
002 -        1  1
003 -        0  0
004 -        0  0
005 -       10  /
006 -       24  DEC
007 -        1  1
008 -        0  0
009 -        0  0
010 -        0  0
011 -        0  0
012 -    42  9  RMD
013 -    43 21  RTN
```
However this will disagree with Excel for 6423 h as the result is 100 d and not 0 (or rather "00").

Now I'm curious what this program is supposed to do. And then I guess the most difficult part will be to produce three question marks.

Cheers

Thomas

Edited: 3 June 2013, 4:06 p.m.

Quote:
that's not what Leonid wants

I stand corrected. If the leading digit is one then the number of immediately following zeroes determine the number of digits of the result.

Thomas, as for the discrepancies in your previous table: for ae01, strip the two rightmost digits to get ae, convert to decimal to get 174, which starts with "1" so retain the rightmost four digits 0174 or 174. For f02, strip to get f, convert to get 15, which again starts with "1" so the rightmost four digits gives 0015 or 15. In neither case does the "1" get removed.

I think I understand the Excel formula and I know that both of my programs don't give the same result in all cases. But I still wonder why 3E800 h, shifted and converted to 1000 d should result in 0 d while A00 h, shifted and converted to 10 d rests as it is. Does that make sense to you?

Kind regards

Thomas

Edited: 3 June 2013, 4:37 p.m.

Quote:
But I still wonder why 3E800 h, shifted and converted to 1000 d should result in 0 d while A00 h, shifted and converted to 10 d rests as it is. Does that make sense to you?

1000, because it begins "1000", is reduced to the rightmost digit to give "0". 10, because it begins "10", is "reduced" to the rightmost three digits to give "010" or "10".

It makes sense according to the rules. I don't see the sense or reason behind the rules, however. Perhaps Leonid could explain what this is all in pursuit of.

Thank you, short program is good solution for me

(Sorry I have not completely described, but you have correctly guessed conditions. This is a practical problem, and not all the possible numbers will be used in it)

By the way, for causes "???" (3dA01). HP-16c has something visual, like Flag 9 in HP-15c?

Quote:
short program is good solution for me

I'm just glad you don't need the equivalent of the Excel-function because that would have been difficult to implement.

I'm not aware that you could make the HP-16C blink as the HP-15C but you can still set flags 4 or 5 to display either the
carry- or the out-of-range-annunciator.

This program sets the G-annunciator for 3DA01 h:

```001 - 43,22, A  LBL A
002 -        1  1
003 -        0  0
004 -        0  0
005 -       10  /
006 -       24  DEC
007 -        1  1
008 -        0  0
009 -        0  0
010 -        0  0
011 -        0  0
012 -    43  3  x>y
013 -    22  3  GTO 3
014 -    42  9  RMD
015 -    43 21  RTN
016 - 43,22, 0  LBL 0
017 -       30  -
018 -    43  2  x<0
019 -    22  1  GTO 1
020 -    43 36  LSTx
021 -    43  1  x<=y
022 -    22  2  GTO 2
023 -       40  +
024 -    43 21  RTN
025 - 43,22, 1  LBL 1
026 -    43 36  LSTx
027 - 43,22, 2  LBL 2
028 -       40  +
029 -    43 36  LSTx
030 - 43,22, 3  LBL 3
031 -        1  1
032 -        0  0
033 -       10  /
034 -    43 30  x>0
035 -    22  0  GTO 0
036 - 43, 4, 5  SF 5
037 -       33  Rv
038 -    43 21  RTN
```

Cheers

Thomas

Edited: 3 June 2013, 10:22 p.m.

Shift right 8 bits, then subtract 10,000(decimal). I haven't used a 16C so I don't know the details of programming it, but that's the idea.

Dave