Need help converting to ASCII Hex format

[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,