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A conversion question (Integer to BCD-string to byte)

Posted on 2003-12-03
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Last Modified: 2010-04-05
Hi all,

I'm currently stuck on a small problem....I know the solution should not be too hard to come by, but I just can't figure it out....

I'm working on an application that communicates with a piece of hardware through the serial port. The communication's up and running, no problems there. There is one problem though, I need to be able to set the time and date of the hardware, and the hardware expects that in bcd (binary coded decimal). The communication expects bytes.

I've written a conversion function, that takes an integer value, and returns a string, which contains a two-digit bcd code. What I want is to convert that string to a byte, and send that byte to the serialport, without any more calculation, to be sure I don't lose the bcd value.

If anyone can think of a better way, I'd love to hear it, this is getting silly!

In my opinion it should be possible to just create a byte out of the two sets of 4 bits the bcd code is made up of, but I'm creating my own blind spots or something...

Binsky


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Question by:Binsky
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5 Comments
 
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Expert Comment

by:wilmsoft
ID: 9868099
I don't understand yet what you want... So, you've done this:

LowBCD.Text := IntToHex(Display,4);


to get the BCD... Now you need which byte?
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by:Binsky
ID: 9872843
Hi,

Actually, what I've done is this :

function TSetupForm.IntToBCD(Value : integer) : string;
begin
   Result := IntToHex(Value div 10, 1) +IntToHex(Value mod 10, 1);
end;

Now I've got a string containing the BCD translation of any integer, but what I want to have is one byte containing the same info...

Binsky
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Author Comment

by:Binsky
ID: 9902341
Okay,

I've answered the question myself...

The conversion is :

BCD := ((16*(DecimalValue div 10)) + (DecimalValue mod 10));
DEC := ((10*(BCDValue div 16))      + (BCDValue mod 16));

Binsky

(Do I get the points myself?) ;-)
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