Convert from string to GUID?
Robert Oschler
creates. So I created a 'const' item like so:
const
CUSTOM_FILE_FORMAT_ID = ['{D60560E0-9FF2-11D6-BD59-00104B3458C5}'];
type
TCustomFileFormatHeader = packed record
ID: TGUID;
end;
What I don't know how to do is convert the editor generated GUID to the
16-byte binary value that TCustomFileFormatHeader.ID requires so I can
initialize the field. How do I do this? Also what _is_ the data format of
the editor generated GUID (Ctrl-Shift-G)? The format used in GUID's is
unfamilar to me: hex-numbers, separated by dashes, enclosed by curly braces,
then single quotes, then square brackets. What format does that represent,
set of bytes, etc?
thx
Robert Oschler
"Robert Oschler" <Osc...@earthlink.net> wrote in message
news:3d403543_2@dnews...
> I want to use a GUID as an identifier for a file format that my program
> creates. So I created a 'const' item like so:
>
Ok nevermind. The answer was simple. Store only the string part of the
GUID in the 'const' item CUSTOM_FILE_FORMAT_ID and use
StringToGUID(CUSTOM_FILE_FORMAT_ID ) to prep the GUID field.
Guido Gybels
> 16-byte binary value that TCustomFileFormatHeader.ID requires
TGUID is a packed record, declared as follows:
TGUID = packed record
D1: Longword;
D2: Word;
D3: Word;
D4: array[0..7] of Byte;
end;
As you can see, the separators in the string representation of the GUID,
represent the separation between the D1-D4 fields of the TGUID. The record
itself is already a binary value. The string REPRESENTS each field of the
record in hexadecimal format.
You don't need to do the calculation yourself though, Delphi allows you to
declare a constant that uses a string literal to specify the GUID value:
const AnyGUID: TGUID = '{12345678-0A0B-0C0D-9000-000000000000}';
You can then use this constant to refer to the binary GUID in your code.
Guido
--
Guido Gybels
www.optimalcode.com/Guido
John Herbster
> D1: Longword;
> D2: Word;
> D3: Word;
> D4: array[0..7] of Byte;
> end;
How in the world did this ever come to be?
Why not just use "array[0..15] of byte"?
William Meyer
> How in the world did this ever come to be?
> Why not just use "array[0..15] of byte"?
Or array[0..3] of LongWord?
And why does the string presentation not match the binary layout? It looks
like:
TGUID = packed record
D1 : LongWord;
D2 : Word;
D3 : Word;
D4 : Word;
D5 : array[0..5] of Byte;
end;
Or maybe more like D5: TribpleWord; (as if we had one <g>)
Who are those guys. anyway? <g>
--
Bill
Education guarantees neither good sense nor rational thinking.
- Posted with XanaNews -
Robert Oschler
"William Meyer" <wmhm...@earthlink.net> wrote in message
news:3d408970_2@dnews...
> John Herbster wrote:
>
end;
>
> Or maybe more like D5: TribpleWord; (as if we had one <g>)
>
> Who are those guys. anyway? <g>
>
Or how come you can't compare two GUID variables without the use of a
special function IsEqualGUID?
Rudy Velthuis (TeamB)
> John Herbster wrote:
>
> > It's that old big/little-endian problem. <g> --JohnH
>
> Yeah, well.... but still, why does the string representation have a
> different grouping than the record?
Who knows? Who invented them anyway? <g>
--
Rudy Velthuis (TeamB)
"If we knew what it was we were doing, it would not be called research,
would it?" -- Albert Einstein
William Meyer
> Who knows? Who invented them anyway? <g>
Well, there's plenty of info here:
http://www.opengroup.org/dce/info/draft-leach-uuids-guids-01.txt
And some more here:
http://gsraj.tripod.com/com/basic_com/naming_guid.html
The latter reference makes it look as though the different formatting of
string and record is based on the original format specs. (But it still
seems goofy to me <g>.)
William Meyer
> It's that old big/little-endian problem. <g> --JohnH
Yeah, well.... but still, why does the string representation have a
different grouping than the record?
--
John Herbster
> Or array[0..3] of LongWord?
It's that old big/little-endian problem. <g> --JohnH