--- In [email protected], "swzoh" <[EMAIL PROTECTED]> wrote:
>
> > maybe in ByteBlock case:
> > 263 = 272 - 6 - 4 + 1
> > where 6 is the byteblock's overhead, and 4 may be terminating null in
> > unicode string case and among them 1 may be counted again as the
> > (plain) terminating null character.
>
> Sorry, I think I counted wrongly here. I don't know why I counted 4,
> not 2 with the unicode string. Hmm, then, I don't know why it's 263.
> Alan may answer it if he reads this thread.
Alan, I think I got a valid conjecture about it. Try the following:
-----------------------------------------------------------------------
local sString,xString
sString="A"
dll.call("MultiByteToWideChar|ui ui s i x i",;;+
0,0,sString,-1,"xString",;;+
dll.call("MultiByteToWideChar|ui ui s i x i",;;+
0,0,sString,-1,"xString",0))
win.debug(binary.length("xString"),binary.dump("xString"))
quit
-----------------------------------------------------------------------
The length of the byteblock "xString" always return 263. I became
curious why it's not 2*2 or 266:
266 = 272 - 1 - 4 - 1
where 1-byte header, 4-byte size as int, 1-byte eos.
My conjecture is that the header is not 1-byte, but actually 4-byte.
Some kind of Struct Packing arose here, I suppose:
263 = 272 - 4 - 4 - 1
This might be also the cause to not report the correct size of
"xString" buffer, 2*2=4.
What do you think?
Sean
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