Is Parallel Programming Hard, And, If So, What Can You Do About It?

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314 APPENDIX F. ANSWERS TO QUICK QUIZZESAnswer:Suppose that the order was reversed, andthat CPU 0 has just reached line 13 ofsynchronize srcu(), while both CPU 1 and CPU 2start executing another synchronize srcu() each,and CPU 3 starts executing a srcu read lock().Suppose that CPU 1 reaches line 6 ofsynchronize srcu() just before CPU 0 incrementsthe counter on line 13. Most importantly,suppose that CPU 3 executes srcu read lock()out of order with the following SRCU read-sidecritical section, so that it acquires a referenceto some SRCU-protected data structure beforeCPU 0 increments sp->completed, but executesthe srcu read lock() after CPU 0 does thisincrement.Then CPU 0 will not wait for CPU 3 to completeits SRCU read-side critical section before exiting the“while” loop on lines 15-16 and releasing the mutex(remember, the CPU could be reordering the code).NowsupposethatCPU2acquiresthemutexnext,and again increments sp->completed. This CPUwill then have to wait for CPU 3 to exit its SRCUread-side critical section before exiting the loop onlines 15-16 and releasing the mutex. But supposethat CPU 3 again executes out of order, completingthe srcu read unlock() prior to executing a finalreference to the pointer it obtained when enteringthe SRCU read-side critical section.CPU 1 will then acquire the mutex, but see thatthe sp->completed counter has incremented twice,and therefore take the early exit. The caller mightwell free up the element that CPU 3 is still referencing(due to CPU 3’s out-of-order execution).To prevent this perhaps improbable,but entirely possible, scenario, the finalsynchronize sched() must precede the mutexrelease in synchronize srcu().Another approach would be to change to comparisonon line 7 of synchronize srcu() to check forat least three increments of the counter. However,such a change would increase the latency of a “bulkupdate” scenario, where a hash table is being updatedor unloaded using multiple threads. In thecurrent code, the latency of the resulting concurrentsynchronize srcu() calls would take at most twoSRCU grace periods, while with this change, threewould be required.More experience will be required to determinewhich approach is really better. For one thing, theremust first be some use of SRCU with multiple concurrentupdaters.Quick Quiz D.5:Wait a minute! With all those new locks, how doyou avoid deadlock?Answer:Deadlock is avoided by never holding more than oneof the rcu_node structures’ locks at a given time.This algorithm uses two more locks, one to preventCPU hotplug operations from running concurrentlywith grace-period advancement (onofflock) andanother to permit only one CPU at a time fromforcing a quiescent state to end quickly (fqslock).These are subject to a locking hierarchy, so thatfqslock must be acquired before onofflock, whichin turn must be acquired before any of the rcu_nodestructures’ locks.Also, as a practical matter, refusing to ever holdmore than one of the rcu_node locks means that itis unnecessary to track which ones are held. Suchtracking would be painful as well as unnecessary.Quick Quiz D.6:Why stop at a 64-times reduction? Why not go fora few orders of magnitude instead?Answer:RCU works with no problems on systems with afew hundred CPUs, so allowing 64 CPUs to contendon a single lock leaves plenty of headroom. Keep inmind that these locks are acquired quite rarely, aseach CPU will check in about one time per graceperiod, and grace periods extend for milliseconds.Quick Quiz D.7:But I don’t care about McKenney’s lame excusesin the answer to Quick Quiz 2!!! I want to get thenumber of CPUs contending on a single lock downto something reasonable, like sixteen or so!!!Answer:OK, have it your way, then!!! SetCONFIG_RCU_FANOUT=16 and (for NR_CPUS=4096)you will get a three-level hierarchy with with256 rcu_node structures at the lowest level, 16rcu_node structures as intermediate nodes, and asingle root-level rcu_node. The penalty you willpay is that more rcu_node structures will need tobe scanned when checking to see which CPUs needhelp completing their quiescent states (256 insteadof only 64).
F.15. CHAPTER D: READ-COPY UPDATE IMPLEMENTATIONS 315Quick Quiz D.8:OK, so what is the story with the colors?Answer:Data structures analogous to rcu_state (includingrcu_ctrlblk) are yellow, those containingthe bitmaps used to determine when CPUs havechecked in are pink, and the per-CPU rcu_datastructures are blue. The data structures used toconserve energy (such as rcu_dynticks) will becolored green.Quick Quiz D.9:Given such an egregious bug, why does Linux runat all?Answer:Because the Linux kernel contains device driversthat are (relatively) well behaved. Few if any ofthem spin in RCU read-side critical sections forthe many milliseconds that would be required toprovoke this bug. The bug nevertheless does needto be fixed, and this variant of RCU does fix it.Quick Quiz D.10:But doesn’t this state diagram indicate thatdyntick-idle CPUs will get hit with reschedule IPIs?Won’t that wake them up?Answer:No. Keep in mind that RCU is handling groups ofCPUs. One particular group might contain bothdyntick-idle CPUs and CPUs in normal mode thathave somehow managed to avoid passing through aquiescent state. Only the latter group will be senta reschedule IPI; the dyntick-idle CPUs will merelybe marked as being in an extended quiescent state.Quick Quiz D.11:But what happens if a CPU tries to report goingthrough a quiescent state (by clearing its bit) beforethe bit-setting CPU has finished?Answer:There are three cases to consider here:1. A CPU corresponding to a non-yet-initializedleaf rcu_node structure tries to report a quiescentstate. This CPU will see its bit alreadycleared, so will give up on reporting its quiescentstate. Some later quiescent state will servefor the new grace period.2. A CPU corresponding to a leaf rcu_node structurethat is currently being initialized tries toreport a quiescent state. This CPU will see thatthe rcu_node structure’s ->lock is held, so willspin until it is released. But once the lock is released,the rcu_node structure will have beeninitialized, reducing to the following case.3. A CPU corresponding to a leaf rcu_node thathasalreadybeeninitializedtriestoreportaquiescentstate. This CPU will find its bit set, andwill therefore clear it. If it is the last CPU forthat leaf node, it will move up to the next levelof the hierarchy. However, this CPU cannotpossiblybethelastCPUinthesystemtoreporta quiescent state, given that the CPU doing theinitialization cannot yet have checked in.So, in all three cases, the potential race is resolvedcorrectly.Quick Quiz D.12:And what happens if all CPUs try to report goingthrough a quiescent state before the bit-settingCPU has finished, thus ending the new grace periodbefore it starts?Answer:The bit-setting CPU cannot pass through a quiescentstate during initialization, as it has irqsdisabled. Its bits therefore remain non-zero, preventingthe grace period from ending until the datastructure has been fully initialized.Quick Quiz D.13:And what happens if one CPU comes out ofdyntick-idle mode and then passed through aquiescent state just as another CPU notices thatthe first CPU was in dyntick-idle mode? Couldn’tthey both attempt to report a quiescent state atthe same time, resulting in confusion?Answer:They will both attempt to acquire the lock on thesame leaf rcu_node structure. The first one toacquire the lock will report the quiescent state andclear the appropriate bit, and the second one toacquire the lock will see that this bit has alreadybeen cleared.
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Is Parallel Programming Hard, And,
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Contents1 Introduction 11.1 Histori
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CONTENTSv6 Locking 676.1 Staying Al
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CONTENTSviiB Synchronization Primit
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CONTENTSixE.7.1 Introduction to Pre
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PrefaceThe purpose of this book is
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Chapter 1IntroductionParallel progr
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1.2. PARALLEL PROGRAMMING GOALS 3CP
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1.3. ALTERNATIVES TO PARALLEL PROGR
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1.4. WHAT MAKES PARALLEL PROGRAMMIN
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1.5. GUIDE TO THIS BOOK 9other hand
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Chapter 2Hardware and its HabitsMos
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2.1. OVERVIEW 13Therefore, as shown
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16 CHAPTER 2. HARDWARE AND ITS HABI
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18 CHAPTER 2. HARDWARE AND ITS HABI
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20 CHAPTER 3. TOOLS OF THE TRADE1 p
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22 CHAPTER 3. TOOLS OF THE TRADE1 p
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24 CHAPTER 3. TOOLS OF THE TRADE1.1
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28 CHAPTER 3. TOOLS OF THE TRADE
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30 CHAPTER 4. COUNTING1 atomic_t co
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32 CHAPTER 4. COUNTING4.2.3 Eventua
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34 CHAPTER 4. COUNTINGvanish when t
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36 CHAPTER 4. COUNTINGper-thread va
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38 CHAPTER 4. COUNTING1 unsigned lo
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42 CHAPTER 4. COUNTING1 #define THE
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46 CHAPTER 4. COUNTINGReadsAlgorith
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48 CHAPTER 5. PARTITIONING AND SYNC
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68 CHAPTER 6. LOCKING1 int delete(i
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70 CHAPTER 7. DATA OWNERSHIP
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110 CHAPTER 9. APPLYING RCU1 struct
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112 CHAPTER 9. APPLYING RCU
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114 CHAPTER 10. VALIDATION: DEBUGGI
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116 CHAPTER 11. DATA STRUCTURES
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118 CHAPTER 12. ADVANCED SYNCHRONIZ
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138 CHAPTER 13. EASE OF USEFigure 1
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140 CHAPTER 13. EASE OF USE
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142 CHAPTER 14. TIME MANAGEMENT
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144 CHAPTER 15. CONFLICTING VISIONS
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154 APPENDIX A. IMPORTANT QUESTIONS
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156 APPENDIX A. IMPORTANT QUESTIONS
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158 APPENDIX B. SYNCHRONIZATION PRI
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160 APPENDIX B. SYNCHRONIZATION PRI
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162 APPENDIX C. WHY MEMORY BARRIERS
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244 APPENDIX E. FORMAL VERIFICATION
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Page 348 and 349: 336 BIBLIOGRAPHY[But97]USA, March 2
Page 350 and 351: 338 BIBLIOGRAPHY[HMB06][Hol03][HP95
Page 352 and 353: 340 BIBLIOGRAPHY[McK06] Paul E. McK
Page 354 and 355: 342 BIBLIOGRAPHYtor. Software - Pra
Page 356 and 357: 344 BIBLIOGRAPHY[UoC08][VGS08]Berke
Page 358: 346 APPENDIX H. CREDITSH.4 Original
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