操作系统概念(第七版_英文版)ch6

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1、Chapter6:ProcessSynchronizationModule6:ProcessSynchronizationBackgroundTheCritical-SectionProblemPeterson’sSolutionSynchronizationHardwareSemaphoresClassicProblemsofSynchronizationMonitorsSynchronizationExamplesAtomicTransactionsBackgroundConcurrentaccesstoshareddatamayresultindatain

2、consistencyMaintainingdataconsistencyrequiresmechanismstoensuretheorderlyexecutionofcooperatingprocessesSupposethatwewantedtoprovideasolutiontotheconsumer-producerproblemthatfillsallthebuffers.Wecandosobyhavinganintegercountthatkeepstrackofthenumberoffullbuffers.Initially,countissett

3、o0.Itisincrementedbytheproducerafteritproducesanewbufferandisdecrementedbytheconsumerafteritconsumesabuffer.SharedMemoryProducer:/*produceanitemandputinnextProduced*/while(true){while(count==BUFFER_SIZE);//donothingbuffer[in]=nextProduced;in=(in+1)%BUFFER_SIZE;count++;}Consumer:/*con

4、sumetheiteminnextConsumed*/while(true){while(count==0);//donothingnextConsumed=buffer[out];out=(out+1)%BUFFER_SIZE;count--;}RaceConditioncount++couldbeimplementedasregister1=countregister1=register1+1count=register1count--couldbeimplementedasregister2=countregister2=register2-1co

5、unt=register2Considerthisexecutioninterleavingwith“count=5”initially:S0:producerexecuteregister1=count{register1=5}S1:producerexecuteregister1=register1+1{register1=6}S2:consumerexecuteregister2=count{register2=5}S3:consumerexecuteregister2=register2-1{register2=4}S4:producerexec

6、utecount=register1{count=6}S5:consumerexecutecount=register2{count=4}WhataboutthecaseifwereversedtheorderofthestatementsatS4andS5?RaceCondition:theoutcomeoftheexecutiondependsontheparticularorderinwhichtheaccesstakesplace.register1andregister2maybethesamephysicalregister,whatwillhap

7、pen?Critical-SectionProblemCritical-Section:inwhichtheprocessesmaybechangingcommonvariables,updatingatable,writingafile,andsoonWemustdesignaprotocolthattheprocessescanusetocooperate,whenoneprocessisexecutinginitscriticalsection,nootherprocessistobeallowedtoexecuteinthiscriticalsectio

8、n.Entrysecti