Re: Help with broken CB chain
I found the statement in the POP... PERFORM LOCKED OPERATION can be thought of as performing concurrent interlocked updates of multiple operands. However, the instruction does not actually perform any interlocked update, and a serially reusable resource cannot be updated predictably through the use of both PERFORM LOCKED OPERATION and conditional-swapping instructions (CS and CDS). Thanks for your help. -- For IBM-MAIN subscribe / signoff / archive access instructions, send email to lists...@listserv.ua.edu with the message: INFO IBM-MAIN
Re: Help with broken CB chain
PLO is basically an "island" of its own. CDSx, transactional and CSST all mix and match (if I have read the PoOp correctly). I don't have the time to read the code at the level of detail necessary to find this kind of flaw but mixing PLO with CDSG is definitely a recipe for the kind of problem you are seeing. Look into CSST. It solved a problem for me. There is a lot of "locked" power there. Charles -Original Message- From: IBM Mainframe Discussion List [mailto:IBM-MAIN@LISTSERV.UA.EDU] On Behalf Of Greg Dyck Sent: Thursday, October 13, 2016 1:52 PM To: IBM-MAIN@LISTSERV.UA.EDU Subject: Re: Help with broken CB chain You *can* *not* safely intermix the use of CDSG and PLO without using special protocols. This is clearly documented in the POPs. You *can* safely intermix CDSG and transactional execution. -- For IBM-MAIN subscribe / signoff / archive access instructions, send email to lists...@listserv.ua.edu with the message: INFO IBM-MAIN
Re: Help with broken CB chain
You *can* *not* safely intermix the use of CDSG and PLO without using special protocols. This is clearly documented in the POPs. You *can* safely intermix CDSG and transactional execution. Regards, Greg -- For IBM-MAIN subscribe / signoff / archive access instructions, send email to lists...@listserv.ua.edu with the message: INFO IBM-MAIN
Re: Help with broken CB chain
Missed a line of code at the end... Here is the updated code: FIRST DS D LASTDS D DOX378 DS0H LGR6,LAST *C IF FIRST AND LAST EQ 0 *C SET FIRST AND LAST = NEW CB ADDRESS XGR R4,R4 XGR R5,R5 LGR R2,R8(R8: ADDRESS OF CB) LGR R3,R8 CDSG R4,R2,FIRST IF FIRST & LAST 0, STM 2,3,FIRST BC4,ELIFX076 *C ELSE BC8,EDOX378 ELIFX076 DS0H *C IF LAST = LAST (R6) *C SET LAST = POINTER_TO_NEW_ (R8) *C SET NEXT.OF.LAST.ONCHAIN= POINTER_TO_NEW_ (R8) STG R6,OPERAND1(COMPARISON VALUE) STG R8,OPERAND1R (REPLACEMENT VALUE) LAR0,CSDSTG LGR R3,R8 LAR4,LAST LAST IS IN CSA STG R4,OPERAND6 DROP R8 USING CB,R6 LAR4,NEXT NEXT IS IN CSA STG R4,OPERAND4 STG R8,OPERAND3 STG R8,OPERAND5 PLO 0,LAST,0,PL * THE FIRST-OPERAND COMPARISON VALUE AND THE SECOND OPERAND ARE * COMPARED. IF THEY ARE EQUAL, THE FIRST-OPERAND REPLACEMENT VALUE * IS STORED AT THE SECOND-OPERAND LOCATION, AND THE THIRD OPERAND IS * STORED AT THE FOURTH-OPERAND LOCATION. THEN, THE FIFTH OPERAND IS * STORED AT THE SIXTH-OPERAND LOCATION. BNZ DOX378 EDOX378 DS0H -- For IBM-MAIN subscribe / signoff / archive access instructions, send email to lists...@listserv.ua.edu with the message: INFO IBM-MAIN
Help with broken CB chain
I have a control block chain in CSA above the BAR that is updated from multiple address spaces with a first and last pointer. The last entry in the chain has a next = 0. If there are no entries in the chain the first and last pointer = 0. I seem to be adding two entries to the beginning of the chain at the same time because rarely one of the control blocks is never freed and it always has a next pointer of zero. I use the following code to add a control block to the end of the chain (always add to the end of the chain): FIRST DS D LASTDS D DOX378 DS0H LGR6,LAST *C IF FIRST AND LAST EQ 0 *C SET FIRST AND LAST = NEW CB ADDRESS XGR R4,R4 XGR R5,R5 LGR R2,R8(R8: ADDRESS OF CB) LGR R3,R8 CDSG R4,R2,FIRST IF FIRST & LAST 0, STM 2,3,FIRST BC4,ELIFX076 *C ELSE BC8,EDOX378 ELIFX076 DS0H *C IF LAST = LAST (R6) *C SET LAST = POINTER_TO_NEW_ (R8) *C SET NEXT.OF.LAST.ONCHAIN= POINTER_TO_NEW_ (R8) STG R6,OPERAND1(COMPARISON VALUE) STG R8,OPERAND1R (REPLACEMENT VALUE) LAR0,CSDSTG LGR R3,R8 LAR4,LAST LAST IS IN CSA STG R4,OPERAND6 DROP R8 USING CB,R6 LAR4,NEXT NEXT IS IN CSA STG R4,OPERAND4 STG R8,OPERAND3 STG R8,OPERAND5 PLO 0,LAST,0,PL * THE FIRST-OPERAND COMPARISON VALUE AND THE SECOND OPERAND ARE * COMPARED. IF THEY ARE EQUAL, THE FIRST-OPERAND REPLACEMENT VALUE * IS STORED AT THE SECOND-OPERAND LOCATION, AND THE THIRD OPERAND IS * STORED AT THE FOURTH-OPERAND LOCATION. THEN, THE FIFTH OPERAND IS * STORED AT THE SIXTH-OPERAND LOCATION. EDOX378 DS 0h Note: I think I could use a CSSTG PLO but this should work and I don’t want to break what normally works since this is production code. I have looked at this code until my eyes are falling out. Does anyone see a flaw in this logic? -- For IBM-MAIN subscribe / signoff / archive access instructions, send email to lists...@listserv.ua.edu with the message: INFO IBM-MAIN
