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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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