No twelve is correct. It contains the entry address. It appears that backing up a little further: 000751B2 000001F2 LA R0,4 4100 0004
000751B6 000001F6 L R15,12(R15,) 58FF 000C 000751BA 000001FA BALR R14,R15 05EF 000751BC 000001FC LTR R15,R15 12FF 000751BE 000001FE BNZ 496(,R12) 4770 C1F0 000751C2 00000202 ST R1,288(,R4) 5010 4120 000751C6 00000206 LA R1,82 4110 0052 000751CA 0000020A AL R1,296(,R4) 5E10 4128 000751CE 0000020E AL R1,992(,R12) 5E10 C3E0 A 12 x'0c' was returned from the BALR which is into an LE module - CEEBPIRA So, until I get the SVC dump, I search on that in the FM, etc. Dave Gibney [EMAIL PROTECTED] System Programmer (509) 335-7359 Information Technology Washington State University Pullman, WA 99164-1222 > -----Original Message----- > From: IBM Mainframe Discussion List [mailto:[EMAIL PROTECTED] On > Behalf Of McKown, John > Sent: Thursday, December 14, 2006 1:32 PM > To: [email protected] > Subject: Re: S0C1 with ILC 6 > > > -----Original Message----- > > From: IBM Mainframe Discussion List > > [mailto:[EMAIL PROTECTED] On Behalf Of Larry Crilley > > Sent: Thursday, December 14, 2006 3:27 PM > > To: [email protected] > > Subject: Re: S0C1 with ILC 6 > > > > > > Well, there you go. > > > > R12 is your base for the BNZ 496(,R12). > > > > Adding the value in R12 to 496 (x'1FE0') yields: 78FD0 + 1F0 = 791C0. > > > > So you would branch there. The code at 791C0 is > > C1F050104120. C1 would be > > ILC 6. > > > > You get this abend when you take the BNZ. > > > > > > Larry Crilley > > IOW, register 12 most likely got corrupted by the previously called > routine? > > -- > John McKown > Senior Systems Programmer > HealthMarkets > Keeping the Promise of Affordable Coverage > Administrative Services Group > Information Technology ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to [EMAIL PROTECTED] with the message: GET IBM-MAIN INFO Search the archives at http://bama.ua.edu/archives/ibm-main.html
