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I think the difference is that matinvat() is a separate function (in *srvpgm
QSYS/QC2UTIL1) so it gets its own invocation stack entry.  But _MATINVAT2 is
inline code that executes in the test() function.

So my answer would be "working as designed".

--Dave

On Monday 26 August 2002 07:12 am, Gene_Gaunt@ReviewWorks.com wrote:
> I noticed the source invocation offset behaves differently between
> _MATINVAT2 and matinvat().  With _MATINVAT2, the source invocation offset
> value '0' denotes the invocation executing _MATINVAT2, value -1 denotes its
> caller, value -2 denotes its caller's caller, etc.  For example the
> following gets the program pointer of the caller, so the source invocation
> offset is -1.
>
> #include <mimchobs.h>
> #include <string.h>
> void test(void)
> { _INV_Template_T  I;
>   _Select_Templatge_T  S;
>   char  *P;
>   memset(&I, 0 ,sizeof(I));
>   memset(&S, 0 ,sizeof(S));
>   I.Inv_Offset = -1;  /*caller*/
>   S.Numb_Attrs = 1;
>   S.Entry[0].Attr_ID = _MATV_PGM_PTR;
>   S.Entry[0].Rcvr_Len = sizeof(P);
>   _MATINVAT2((_SPCPTR) &P, &I, &S);
>   return; }
>
> But with matinvat(), the source invocation offset value -1 denotes the
> invocation executing matinvat(), value -2 denotes its caller, value -3
> denotes its caller's caller, etc.  So to acheive the same result above with
> matinvat(), I have to decrement the offset from -1 to -2!  Is this working
> as designed, or an off by one bug?
>
> #include <mimchobs.h>
> #include <string.h>
> void test(void)
> { _INV_Template_T  I;
>   char  *P;
>   memset(&I, 0, sizeof(I));
>   I.Inv_Offset = -2;   /*caller*/
>   matinvat((_SPCPTR) &P, &I, _MATV_PGM_TR, sizeof(P));
>   return; }



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