Classical%20Synchronization%20Problems

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Classical Synchronization Problems

• CS 537 - Introduction to Operating Systems

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Classical Problem 1

• Producer-Consumer

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• include ltstdio.hgt
• include ltsemaphore.hgt
• define SIZE 50
• define NUM_THREADS 5
• struct object buffer
• sem_t mutex, empty, full
• void nap()
• int sleepTime (int)(5 random())
• sleep(sleepTime)
• void enter(struct object item)
• empty.down()
• mutex.down()
• count
• bufferin item

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• void producer()
• struct object newObj
• for()
• nap()
• newObj getObject()
• enter(newObj)
• void consumer()
• struct object item
• for()
• nap()
• item remove()
• useItem(item)
• int main(int argc, char argv)

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Classical Problem 2

• Readers-Writers

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Readers-Writers

• Several threads accessing a database
• Some threads read the database (readers)
• Some threads write the database (writers)
• Multiple readers can be accessing database at a
  time if no writers accessing it
• Only one thread can be accessing database if it
  is a writer
• 2 major variations
• no reader waits if no writer in database
• writers can starve
• all readers wait if a writer is waiting
• writer has to wait until current reader (or
  writer) is finished with database
• readers may starve
• Many more variations to deal with starvation
  issues
• Following example is for the first variant
• starvation is disregarded

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• include ltstdio.hgt
• include ltpthreads.hgt
• define NUM_THREADS 5
• sem_t mutex, database
• void reader()
• while(true)
• nap()
• startRead()
• nap()
• endRead()
• void writer()
• while(true)
• nap()

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• void startRead()
• mutex.down()
• readerCount
• if(readerCount 1) database.down()
• mutex.up()
• void endRead()
• mutex.down()
• readerCount--
• if(readerCount 0) database.up()
• mutex.up()
• void startWrite() database.down()
• void endWrite() database.up()
• int main()
• pthread_t threadsNUM_THREADS