structure incompatibility between VB and VC++

M

Mohamed Fysal

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..........
 
T

tragomaskhalos

[snip]
Click to expand...

You don't really give enough info, but VB strings are
COM BSTRs, which store the length in the first two bytes.
I suspect this is the root of you problem, but try an
on-topic NG or google for help.
 
J

Jim Langston

Mohamed Fysal said:
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..........
Click to expand...

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.
 
C

Cholo Lennon

Hi,
Click to expand...
I have a major problem.
Click to expand...
I am using a VB exe which is having a structure declared and it looks like
below
Click to expand...
Public Type SCANNEDIBUTTNO
IButtonUserNumber As String * 32
End Type
Click to expand...
Public Type BOOKINGIDPARAM
bkidno(14) As SCANNEDIBUTTNO
End Type
Click to expand...
Public Type TRACKERMESSAGEINFO
bkidp As BOOKINGIDPARAM
End Type
Click to expand...
And I have declared a varaible for TRACKERMESSAGEINFO as tmi
Click to expand...
and the value for IButtonUserNumber is a Hex value which is stored as
follows:
Click to expand...
textValue1 = 'EFABCD12'
textValue2 = 'EFABCD13'
textValue3 = 'EFABCD14'
Click to expand...
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.....
Click to expand...
When the above structure variable tmi is passed to a VC++ Win 32 DLL, the
string value is stored in a different way.
Click to expand...
whereby I have declared the structure in VC++ DLL as follows:
Click to expand...
typedef struct tagSCANNEDIBUTTONNO
{
char IButtonUserNumber[32];
} SCANNEDIBUTTNO, FAR * LPSCANNEDIBUTTNO;
typedef struct tagBOOKINGIDPARAM
{
SCANNEDIBUTTNO bkidno[15];
} BOOKINGIDPARAM, FAR * LPBOOKINGIDPARAM;
Click to expand...
typedef struct tagTRACKERMESSAGEINFO
{
BOOKINGIDPARAM bkidp;
} TRACKERMESSAGEINFO, FAR * LPTRACKERMESSAGEINFO;
Click to expand...
When the value is passed to the VC++ dll i get the value as follows
Click to expand...
LPTRACKERMESSAGEINFO lptmi
Click to expand...
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
Click to expand...

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
Click to expand...

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] = ''
Click to expand...
So i do not know why there is an empty space of two bytes before the
actual values (EFABCD12) , (EFABCD13) , and (EFABCD14) ?????????????
Click to expand...
Please help me in this regard..........
Click to expand...

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.
Click to expand...

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.


Regards
 
J

Jim Langston

Cholo Lennon said:
Hi,
Click to expand...
I have a major problem.
Click to expand...
I am using a VB exe which is having a structure declared and it looks
like
below
Click to expand...
Public Type SCANNEDIBUTTNO
IButtonUserNumber As String * 32
End Type
Click to expand...
Public Type BOOKINGIDPARAM
bkidno(14) As SCANNEDIBUTTNO
End Type
Click to expand...
Public Type TRACKERMESSAGEINFO
bkidp As BOOKINGIDPARAM
End Type
Click to expand...
And I have declared a varaible for TRACKERMESSAGEINFO as tmi
Click to expand...
and the value for IButtonUserNumber is a Hex value which is stored as
follows:
Click to expand...
textValue1 = 'EFABCD12'
textValue2 = 'EFABCD13'
textValue3 = 'EFABCD14'
Click to expand...
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.....
Click to expand...
When the above structure variable tmi is passed to a VC++ Win 32 DLL,
the
string value is stored in a different way.
Click to expand...
whereby I have declared the structure in VC++ DLL as follows:
Click to expand...
typedef struct tagSCANNEDIBUTTONNO
{
char IButtonUserNumber[32];
} SCANNEDIBUTTNO, FAR * LPSCANNEDIBUTTNO;
typedef struct tagBOOKINGIDPARAM
{
SCANNEDIBUTTNO bkidno[15];
} BOOKINGIDPARAM, FAR * LPBOOKINGIDPARAM;
Click to expand...
typedef struct tagTRACKERMESSAGEINFO
{
BOOKINGIDPARAM bkidp;
} TRACKERMESSAGEINFO, FAR * LPTRACKERMESSAGEINFO;
Click to expand...
When the value is passed to the VC++ dll i get the value as follows
Click to expand...
LPTRACKERMESSAGEINFO lptmi
Click to expand...
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
Click to expand...

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
Click to expand...

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] = ''
Click to expand...
So i do not know why there is an empty space of two bytes before the
actual values (EFABCD12) , (EFABCD13) , and (EFABCD14) ?????????????
Click to expand...
Please help me in this regard..........
Click to expand...

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.
Click to expand...

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.
Click to expand...

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.
 
C

Cholo Lennon

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.
Click to expand...
Click to expand...
Correction: BSTR string are null terminated.
Click to expand...
Check this for more details:
http://blogs.msdn.com/ericlippert/archive/2003/09/12/52976.aspx
Click to expand...
Be warned that BSTR strings are wide strings (2 bytes per character).
That would be the reason of the problem.
Click to expand...

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=16...

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.
Click to expand...

I never said that BSTR string hasn't its length at the front. I only
said that they are NULL terminated (to provide C/Win API
compatibility).
You can see the memory layout of BSTR strings in the chapter 2, topic
"Dealing with String / Strings Inside Out" from the book "Hardcore
Visual Basic" (Bruce McKinney 1997). Online version at http://vb.mvps.org/hardcore/

Best Regards
 

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