Need help converting to ASCII Hex format

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- 25th August 2007, 19:20
tcbcats

Need help converting to ASCII Hex format

This is the task…
I have a number stored in a 16 bit word. The number is up to 5 digits like 01389.
I can also convert it to the format $30, $31, $33, $38, $39 in a 4 byte string. If needed. Each digit is 0 to 9 decimal… never any alpha chr’s. Max value for the expected data is 09999

I am looking to convert the 5-digit number into a 3 byte “ASCII Hex format” so the number above would look like… $00, $13, $89 in 3 hex bytes. (Looks like the decimal number when looking at the 3 byte HEX code.)

The first byte is not used in this application so it can be hard coded to $00
I need some help on how to do this.

Thanks,
Tcbcats
- 25th August 2007, 19:54
Robson

Where is your data stored ?
Variable or EEprom or something else?
- 25th August 2007, 20:32
BobP

Hi tcbcats,

This is just off the top of my head and will need written and debugged.

Quote:

I can also convert it to the format $30, $31, $33, $38, $39 in a 4 byte string. If needed. Each digit is 0 to 9 decimal… never any alpha chr’s. Max value for the expected data is 09999

Subtract each number with $30 giving 00, 01, 03, 08 and 09 Then

For example to get the first LSB = (08*16)+09 this should give $89
Do the same for the other numbers.

Please this is off the top of my head and I havn't checked it out! It may be of some help?

Bob
- 25th August 2007, 21:07
tcbcats

Stored in a variable

Quote:

Originally Posted by Robson

Where is your data stored ?
Variable or EEprom or something else?

It is in a variable... not in EEprom or from a serial source.
Tcbcats
- 25th August 2007, 22:14
Robson

Try this

Code:

B0 VAR BYTE B1 VAR BYTE B2 VAR BYTE B3 VAR BYTE BWord1 Var BYTE BWord2 VAR BYTE BWord VAR WORD VarWord VAR WORD VarWord = 1389 ' Here is your number you defined B3 = VarWord DIG 3 B2 = VarWord DIG 2 B1 = VarWord DIG 1 B0 = VarWord DIG 0 BWord1 = B3 * 16 + B2 BWord2 = B1 * 16 + B0 BWord = BWord1 * $100 + BWord2 ' Here stands your Word like 1389 as hex
- 25th August 2007, 22:53
Darrel Taylor

Code:

Value VAR WORD Buf VAR BYTE[3] DigLoop VAR BYTE Value = 01389 For DigLoop = 5 to 1 STEP -2 Buf((6-DigLoop)/2) = ((Value DIG DigLoop) << 4) + Value DIG (DigLoop -1) Next DigLoop LCDOUT HEX2 Buf(0),":",HEX2 Buf(1),":",HEX2 Buf(2)

Displays 00:13:89
<br>
- 25th August 2007, 23:49
Robson

Darrel something is not going in my head.

Code:

Value = 01389 ' <- Is the "0" really included ?
- 26th August 2007, 00:17
Darrel Taylor

The compiler will ignore any leading 0's.
That's the way Tcbcats showed it, so I left it in there.
<br>
- 26th August 2007, 11:43
Robson

@Darrel
i mean this line.

Code:

For DigLoop = 5 to 1 STEP -2 Buf((6-DigLoop)/2) ' = Buf((6-5)/2) = 0.5 ??? Does it really works? Round it to zero? Next DigLoop

I don´t tried it out, maybe it works
- 26th August 2007, 19:14
Darrel Taylor

Sure it works. :)

But PBP uses integer math. So there are no decimals.

With STEP -2 the loop will execute 3 times, with the values 5, 3, 1 in DigLoop.
Since it's Integer math, (6-DigLoop)/2 will return 0, 1 and 2 respectively.

Code:

loop Buf((6-DigLoop)/2) = ((Value DIG DigLoop) << 4) + Value DIG (DigLoop -1) 5 BUF( 0 ) = ( $00 ) + 0 = $00 3 BUF( 1 ) = ( $10 ) + 3 = $13 1 BUF( 2 ) = ( $80 ) + 9 = $89

hth,