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Chapter6:ProcessSynchronization
Module6:ProcessSynchronizationBackgroundTheCritical-SectionProblemPeterson’sSolutionSynchronizationHardwareSemaphoresClassicProblemsofSynchronizationMonitorsSynchronizationExamplesAtomicTransactions
BackgroundConcurrentaccesstoshareddatamayresultindatainconsistencyMaintainingdataconsistencyrequiresmechanismstoensuretheorderlyexecutionofcooperatingprocessesSupposethatwewantedtoprovideasolutiontotheconsumer-producerproblemthatfillsallthebuffers.Wecandosobyhavinganintegercountthatkeepstrackofthenumberoffullbuffers.Initially,countissetto0.Itisincrementedbytheproducerafteritproducesanewbufferandisdecrementedbytheconsumerafteritconsumesabuffer.
SharedMemoryProducer:/*produceanitemandputinnextProduced*/while(true){ while(count==BUFFER_SIZE) ;//donothing buffer[in]=nextProduced;in=(in+1)%BUFFER_SIZE;count++;}Consumer:/*consumetheiteminnextConsumed*/while(true){ while(count==0) ;//donothing nextConsumed=buffer[out];out=(out+1)%BUFFER_SIZE; count--; }
RaceConditioncount++couldbeimplementedas
register1=count
register1=register1+1
count=register1count--couldbeimplementedas
register2=count
register2=register2-1
count=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:producerexecutecount=register1{count=6}
S5:consumerexecutecount=register2{count=4}WhataboutthecaseifwereversedtheorderofthestatementsatS4andS5?RaceCondition:theoutcomeoftheexecutiondependsonthepar
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