Yeah, I suppose a routine could be looking at R14 minus 10 to see how it got there. Unlikely, and awful coding technique, but possible. But why only this one routine? If I comment out the call in the (large, real) C++ program, a subsequent call to another routine in the same source module works.
I would assume C++ gets the stack at startup, not on the first external call. Interesting thought. I could add a call to a dummy do-nothing routine ahead of the ISAUTH call. But frankly, I am about out of patience with this problem. I have a fix now for the "real" problem, so perhaps I need to get back to my real job. Charles -----Original Message----- From: IBM Mainframe Discussion List [mailto:IBM-MAIN@LISTSERV.UA.EDU] On Behalf Of Bernd Oppolzer Sent: Thursday, May 28, 2015 7:21 AM To: IBM-MAIN@LISTSERV.UA.EDU Subject: Re: Mysterious U4088-63 from RPTSTG(ON) My only guess is: the branch in the Prolog to obtain New storage in the case when the current segment is too small must point to a special routine in the rptstg(on) case. And maybe this routine has a Problem if the branch is a BRC and no BC. But I cannot really imagine what sort of problem ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to lists...@listserv.ua.edu with the message: INFO IBM-MAIN
