Need help converting to ASCII Hex format

[tcbcats]

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

[Robson]

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

[BobP]

Hi tcbcats,

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


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

[tcbcats]

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

[Robson]

Try this


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

[Darrel Taylor]

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>

[Robson]

Darrel something is not going in my head.


Value = 01389 ' <- Is the "0" really included ?

[Darrel Taylor]

The compiler will ignore any leading 0's.
That's the way Tcbcats showed it, so I left it in there.
<br>

[Robson]

@Darrel
i mean this line.



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

[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.



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,