Ok,
Thank you so much!i'll follow your suggestion
Regards
Samuele Gazzani
Torsten Levin wrote:
> CVI (and every other compiler) packs structures to boundaries depending
> on the processor type and operating system. On windows the default
> packing is ( i guess) on 4 byte boundaries. That's why your short
> occupies 4 bytes rather than 2. The same will happen if you define a
> byte value (char) in a structure, then there are 3 bytes overhead.
>
> CVI offers the possibility to control the packing of data structures by
> pre-compiler directives:
>
> #pragma pack(push) // save the default packing (normally 4)
> #pragma pack(1) // set new packing to 1 byte
>
> typedef struct // define your "packed" struct
> {
> short var1;
> char
var2;
> ....
> ....
> } DataType1; DataType1 data1;
>
> #pragma pack(pop) // restore the previous packing
>
> Regards,
> Torsten
>
> samuele gazzani schrieb:
> >
> > Hi,
> > I found this problem using a "short type" inside a structure: in the
> > LabWindows/CVI Programmer Reference Manual page 1-5 the type "short" is
> > declared to have a size of 16 (2 bytes) . I used the function sizeof()
> > to get the size of a variable short and it returned 2; but when I tried
> > to use the same function for a member of a structure of type short, it
> > returned 4! How can this happen? A short is a short...am I wrong?
> > Example
> > short var1;
> >
> > typedef struct
> > {
> > short var1;
> > ....
> > ....
> > } DataType1; DataType1 data1;
> >
> > sizeof(var1) : returned 2
> > sizeof(data1.var1) : returned 4 ......mistery of my compiler?
> >
> > Thanks a lot
> >
> > Samuele Gazzani
>
> --
> Torsten Levin Tel.: ++49-(0)89-72495-451
> Kayser-Threde GmbH
Fax: ++49-(0)89-72495-291
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