Hexadecimal to Decimal conversion without 2's complement

[therabbi99]

Hello,

I'm brand new to DIAdem and attempting to use the calculator to convert a channel from hexadecimal values to multiple individual binary bit channels. 

I have a formula that works and formats everything the way I want, but the step going from Hex to Dec seems to be defaulting to 2's complement.  However, I need an unsigned decimal.  The formula I'm using for this step in the conversion is below:

ch("[1]/decarr")=CLng("&h" & ch("[1]/hexarr"))

So for example, I have a hex string of 88288900

When I apply the above formula I get -2010609408

However, I'm looking for 2284357888

Is there a simple way to achieve this?

Thank you in advance!

Jeff

[AndreasK · Accepted solution]

As long as I know there is no Unsigned 32 bit type in vbs.

So if you need it to be unsigned you need to split it using Left and Right.

MsgBox Hex(CLng("&h88288900")) & " " & CLng("&h88288900")
MsgBox Hex("&h" & Left("88288900", 4)) & " " & CLng("&h" & Left("88288900", 4)) & VBCRLF &_
       Hex("&h" & Right("88288900", 4)) & " " & CLng("&h" & Right("88288900", 4))

[therabbi99]

Thank you AndreasK!

I did as you suggested.  Unfortunately, I don't have access to write scripts.  So I'm doing everything using the Calculator.  For the benefit of others, here is what I ended up with:

Removes "0x" characters from front end of hexadecimal (if required)
ch("[1]/hex")=MID(INSERT CHANNEL,3)

If NOT using two's complement:

All functions below combined into one. Creates a channel for Bit number 12.
ch("[1]/Bit 12")=CDbl(Mid(Str(Right("0000000000000000" & Str(CLng("&h" & Left(ch("[1]/hex"), 4)),"b"),16)) & Str(Right("0000000000000000" & Str(CLng("&h" & Right(ch("[1]/hex"), 4)),"b"),16)),(33-12),1))

Splits hexadecimal and converts each half to decimal
ch("[1]/dec1")=CLng("&h" & Left(ch("[1]/hex"), 4))
ch("[1]/dec2")=CLng("&h" & Right(ch("[1]/hex"), 4))

Converts each half to binary
ch("[1]/bin1")=Str(ch("[1]/dec1"),"b")
ch("[1]/bin2")=Str(ch("[1]/dec2"),"b")

Pads front end of binary in order to get 16 bits
ch("[1]/bin1padded")=Right("0000000000000000" & ch("[1]/bin1"),16)
ch("[1]/bin2padded")=Right("0000000000000000" & ch("[1]/bin2"),16)

Combines 2 binary strings
ch("[1]/bin")=Str(ch("[1]/bin1padded")) & Str(ch("[1]/bin2padded"))

Creates a channel for Bit number 12
ch("[1]/Bit 12")=CDbl(Mid(ch("[1]/bin"),(33-12),1))

Cheers!