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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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