操作系统概念（第七版英文版）.ppt

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1、Chapter6:ProcessSynchronizationModule6:ProcessSynchronizationBackgroundTheCritical-SectionProblemPeterson’sSolutionSynchronizationHardwareSemaphoresClassicProblemsofSynchronizationMonitorsSynchronizationExamplesAtomicTransactionsBackgroundConcurrentaccesstos

2、hareddatamayresultindatainconsistencyMaintainingdataconsistencyrequiresmechanismstoensuretheorderlyexecutionofcooperatingprocessesSupposethatwewantedtoprovideasolutiontotheconsumer-producerproblemthatfillsallthebuffers.Wecandosobyhavinganintegercountthatkeep

3、strackofthenumberoffullbuffers.Initially,countissetto0.Itisincrementedbytheproducerafteritproducesanewbufferandisdecrementedbytheconsumerafteritconsumesabuffer.SharedMemoryProducer:/*produceanitemandputinnextProduced*/while(true){while(count==BUFFER_SIZE);//

4、donothingbuffer[in]=nextProduced;in=(in+1)%BUFFER_SIZE;count++;}Consumer:/*consumetheiteminnextConsumed*/while(true){while(count==0);//donothingnextConsumed=buffer[out];out=(out+1)%BUFFER_SIZE;count--;}RaceConditioncount++couldbeimplementedasregister1=count

5、register1=register1+1count=register1count--couldbeimplementedasregister2=countregister2=register2-1count=register2Considerthisexecutioninterleavingwith“count=5”initially:S0:producerexecuteregister1=count{register1=5}S1:producerexecuteregister1=register1+

6、1{register1=6}S2:consumerexecuteregister2=count{register2=5}S3:consumerexecuteregister2=register2-1{register2=4}S4:producerexecutecount=register1{count=6}S5:consumerexecutecount=register2{count=4}WhataboutthecaseifwereversedtheorderofthestatementsatS4and

7、S5?RaceCondition:theoutcomeoftheexecutiondependsontheparticularorderinwhichtheaccesstakesplace.register1andregister2maybethesamephysicalregister,whatwillhappen?Critical-SectionProblemCritical-Section:inwhichtheprocessesmaybechangingcommonvariables,updatingat

8、able,writingafile,andsoonWemustdesignaprotocolthattheprocessescanusetocooperate,whenoneprocessisexecutinginitscriticalsection,nootherprocessistobeallowedtoexecuteinthiscriticalsection.Entrysecti