verifying a two-lock concurrent queue hussain tinwala fall 2007

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Reviewing the Simulation

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Verifying a Two-LockConcurrent Queue

Hussain TinwalaFall 2007

• Simulation Review• Searching for Deadlocks• Race Conditions• Queue Verification

Overview

Reviewing the Simulation

Searching for Deadlocks (1)• JPF did not find errors…• Introduced a problem I faced during implementation with

Dequeuer threads:

Searching for Deadlocks (2)

Searching for Deadlocks (3)Run Simulation with 1 Enqueuer, 1 Dequeuerand 1 node:jpf -c jpf.properties…

gov.nasa.jpf.jvm.choice.ThreadChoiceFromSet {Thread-0,>Thread-1} Enqueuer.java:57 : synchronized(sim) Enqueuer.java:59 : sim.notifyAll();------------------------------------------------------ transition #47 thread: 1gov.nasa.jpf.jvm.choice.ThreadChoiceFromSet {>Thread-0} Dequeuer.java:38 : synchronized (sim) Dequeuer.java:40 : sim.wait();====================================================== snapshot #1thread index=1,name=Thread-0,status=WAITING,this=Dequeuer@246,priority=5,lockCount=1 waiting on: Simulator@232 call stack:at Dequeuer.run(Dequeuer.java:40)====================================================== resultserror #1: gov.nasa.jpf.jvm.NotDeadlockedProperty \"deadlock encountered: thread index=0,name=main,s..."

Race Conditions (1)Two types:

- Critical: different outcomes- Non-critical: eventual outcome of the program is the same

Both aresynchronized methods

modifyread

Race Conditions (2)

prompt> jpf -c jpf.properties …- Removed `synchronized’

- jpf with 1 Enqueuer and 1 Dequeuer --> no error- jpf with 1 Enqueuer and n Dequeuers --> no error- jpf with >1 Enqueuer --> error

- Enqueuer threads use both methods- modifies a value- reads a value

- Dequeuer threads only use `getNumEnqsLeft’- reads a value

- Potential for a race condition…

Race Conditions (3)- Using JPF to detect race conditions:

- Add to configuration file:# listenersjpf.listener=gov.nasa.jpf.tools.precise.PreciseRaceDetector

- Running JPF with 2 Enqueuers and 1 Dequeuer- jpf -c jpf.properties…

name=Thread-1,status=RUNNING,this=Enqueuer@136,priority=5,lockCount=0at Simulator.getNumEnqsLeft(Simulator.java:57)at Enqueuer.run(Enqueuer.java:52)

Name=Thread-2,status=RUNNING,this=Enqueuer@137,priority=5,lockCount=0 owned locks:Simulator@232 (because method is synchronized)

at Simulator.decrementNumEnqsLeft(Simulator.java:53)at Enqueuer.run(Enqueuer.java:51)

=========================================== resultserror #1: gov.nasa.jpf.tools.PreciseRaceDetector "race for: "int Simulator.numEnqsLeft" Thread-1 and Thread-2...”=========================================== resultserror #1: RaceDetector "potential field race: Simulator@232.numEnqsLeft"

So we have 2 Enqueuers - One is updating the value - The other is attempting to read it

Race Conditions (4)- Adding `synchronized’ to the methods removes the race condition- Question: Do both methods need to be synchronized?

- Yes: To get rid of the race condition- No: A race does not have to be bad

- Synchronizing the method that modifies a value may be sufficient

- Added `synchronized’ only to the decrementNumEnqsLeft method which is used by Enqueuer threads.

- getNumEnqsLeft method left without `synchronized’- Tested again: jpf -c jpf.properties …

Race Conditions (5)- JPF finds no deadlocks or errors but it does find a race condition

- But that is okay… it is a non-critical race condition

- No matter what order the execution takes, there will always be AT LEAST one Enqueuer thread that will find

- getNumEnqsLeft() == 0- And so, it will be able to notify any waiting Dequeuer threads.

- Overall Advantage: There is less blocking in the Enqueuer thread

Verifying the Queue• Nodes are always inserted after the last node in the

linked list• Nodes are always deleted from the beginning of the

linked list• Head always points to the first node in the list• Tail always points to the last node in the list

• Potential Approach:– Use a PropertyListener

• Ex: QueuePropertyListener extends PropertyListenerAdapter– Then at each transition, verify the above mentioned

properties

Thank you

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