Re: structure incompatibility between VB and VC++

From:

"Jim Langston" <tazmaster@rocketmail.com>

Newsgroups:

comp.lang.c++

Date:

Sat, 10 Nov 2007 06:50:01 -0800

Message-ID:

<SwjZi.15$1z4.14@newsfe06.lga>

"Cholo Lennon" <chololennon@hotmail.com> wrote in message
news:1194639444.289970.143560@22g2000hsm.googlegroups.com...

On Nov 9, 3:25 pm, "Jim Langston" <tazmas...@rocketmail.com> wrote:

"Mohamed Fysal" <fy...@bokia.net> wrote in message

news:4733cca7_2@news.tm.net.my...

Hi,

I have a major problem.

I am using a VB exe which is having a structure declared and it looks
like
below

Public Type SCANNEDIBUTTNO
   IButtonUserNumber As String * 32
End Type

Public Type BOOKINGIDPARAM
    bkidno(14) As SCANNEDIBUTTNO
End Type

Public Type TRACKERMESSAGEINFO
   bkidp As BOOKINGIDPARAM
End Type

And I have declared a varaible for TRACKERMESSAGEINFO as tmi

and the value for IButtonUserNumber is a Hex value which is stored as
follows:

textValue1 = 'EFABCD12'
textValue2 = 'EFABCD13'
textValue3 = 'EFABCD14'

Mid$(tmi.bkidp.bkidno(0).IButtonUserNumber, 1, Len(textValue1)) =
textValue1
Mid$(tmi.bkidp.bkidno(1).IButtonUserNumber, 1, Len(textValue2)) =
textValue2
Mid$(tmi.bkidp.bkidno(2).IButtonUserNumber, 1, Len(textValue3)) =
textValue3
and so on.....

When the above structure variable tmi is passed to a VC++ Win 32 DLL,
the
string value is stored in a different way.

whereby I have declared the structure in VC++ DLL as follows:

typedef struct tagSCANNEDIBUTTONNO
{
   char IButtonUserNumber[32];
} SCANNEDIBUTTNO, FAR * LPSCANNEDIBUTTNO;
typedef struct tagBOOKINGIDPARAM
{
    SCANNEDIBUTTNO bkidno[15];
} BOOKINGIDPARAM, FAR * LPBOOKINGIDPARAM;

typedef struct tagTRACKERMESSAGEINFO
{
    BOOKINGIDPARAM bkidp;
} TRACKERMESSAGEINFO, FAR * LPTRACKERMESSAGEINFO;

When the value is passed to the VC++ dll i get the value as follows

LPTRACKERMESSAGEINFO lptmi

lptmi->bkidp.bkidno(0).IButtonUserNumber[0] = ''
lptmi->bkidp.bkidno(0).IButtonUserNumber[1] = ''
lptmi->bkidp.bkidno(0).IButtonUserNumber[2] = 'E'
lptmi->bkidp.bkidno(0).IButtonUserNumber[3] = 'F'
lptmi->bkidp.bkidno(0).IButtonUserNumber[4] = 'A'
lptmi->bkidp.bkidno(0).IButtonUserNumber[5] = 'B'
lptmi->bkidp.bkidno(0).IButtonUserNumber[6] = 'C'
lptmi->bkidp.bkidno(0).IButtonUserNumber[7] = 'D'
lptmi->bkidp.bkidno(0).IButtonUserNumber[8] = '1'
lptmi->bkidp.bkidno(0).IButtonUserNumber[9] = '2'
lptmi->bkidp.bkidno(0).IButtonUserNumber[10] = ''
lptmi->bkidp.bkidno(0).IButtonUserNumber[11] = ''
...
...
..
lptmi->bkidp.bkidno(0).IButtonUserNumber[31] = ''

and

lptmi->bkidp.bkidno(1).IButtonUserNumber[0] = ''
lptmi->bkidp.bkidno(1).IButtonUserNumber[1] = ''
lptmi->bkidp.bkidno(1).IButtonUserNumber[2] = 'E'
lptmi->bkidp.bkidno(1).IButtonUserNumber[3] = 'F'
lptmi->bkidp.bkidno(1).IButtonUserNumber[4] = 'A'
lptmi->bkidp.bkidno(1).IButtonUserNumber[5] = 'B'
lptmi->bkidp.bkidno(1).IButtonUserNumber[6] = 'C'
lptmi->bkidp.bkidno(1).IButtonUserNumber[7] = 'D'
lptmi->bkidp.bkidno(1).IButtonUserNumber[8] = '1'
lptmi->bkidp.bkidno(1).IButtonUserNumber[9] = '3
lptmi->bkidp.bkidno(1).IButtonUserNumber[10] = ''
lptmi->bkidp.bkidno(1).IButtonUserNumber[11] = ''
...
...
..
lptmi->bkidp.bkidno(1).IButtonUserNumber[31] = ''

and

lptmi->bkidp.bkidno(2).IButtonUserNumber[0] = ''
lptmi->bkidp.bkidno(2).IButtonUserNumber[1] = ''
lptmi->bkidp.bkidno(2).IButtonUserNumber[2] = 'E'
lptmi->bkidp.bkidno(2).IButtonUserNumber[3] = 'F'
lptmi->bkidp.bkidno(2).IButtonUserNumber[4] = 'A'
lptmi->bkidp.bkidno(2).IButtonUserNumber[5] = 'B'
lptmi->bkidp.bkidno(2).IButtonUserNumber[6] = 'C'
lptmi->bkidp.bkidno(2).IButtonUserNumber[7] = 'D'
lptmi->bkidp.bkidno(2).IButtonUserNumber[8] = '1'
lptmi->bkidp.bkidno(2).IButtonUserNumber[9] = '4'
lptmi->bkidp.bkidno(2).IButtonUserNumber[10] = ''
lptmi->bkidp.bkidno(2).IButtonUserNumber[11] = ''
...
...
..
lptmi->bkidp.bkidno(2).IButtonUserNumber[31] = ''

So i do not know why there is an empty space of two bytes before the
actual values (EFABCD12) , (EFABCD13) , and (EFABCD14) ?????????????

Please help me in this regard..........

As Tragomaskhalos says, basic strings use the first two bytes as the
length
of the strings. They are not null terminated as c-style strings are.

Since the length of the strings are 8 one of those bytes is an 8 which is
an
unprintable character which is why you don't see anything. I'm not sure
if
[0] or [1] would be the 8, you'll have to check.

Correction: BSTR string are null terminated.

Check this for more details:
http://blogs.msdn.com/ericlippert/archive/2003/09/12/52976.aspx

Be warned that BSTR strings are wide strings (2 bytes per character).
That would be the reason of the problem.

According to MSDN they contain the length of the string in front of the
characters.
http://www.codeproject.com/string/bstrsproject1.asp?df=100&forumid=16579&exp=0&select=1122506

Quote: "C strings are arrays of characters terminated by a NULL character.
Visual Basic strings differ in that the length of the string preceded the
characters in the string. So a VB string knows it's own length. In
addition, all VB strings are Unicode (16 bits per character)."

Interesting that we can see the 2 bytes for the length of the string in
front, but each character seems to be fitting in one byte.