Synchronization, part 3 Monitors, classical sync. problems Operating Systems Synchronization, part 3 Monitors, classical sync. problems

Monitor Monitor – a synchronization primitive A monitor is a collection of procedures, variables and data structures, grouped together Mutual Exclusion – only one process can be active within a monitor at any given time Usually a programming language construct! The compiler of the language will know that monitors procedures are different than other procedures, and will treat them differently. That means that the compiler is in charge of the mutual exclusion implementation

Condition variables A way for processes to block when they can’t continue Despite its name, it is used to indicate an event and not as a regular valued variable. A CV is not a counter! Two operations: wait, signal Wait: causes the process to block, and allows entry of other threads to the monitor Signal: More than just one alternative: Hoare type monitors: The signaler yields the monitor to the released thread. Signal will be the last operation within the monitor, which wakes up waiting processes (waiting on the same variable). This is not true for Java. Mesa type monitors: The signaling process is allowed to continue http://download.oracle.com/javase/6/docs/api/java/lang/Object.html#notify%28%29 Wakes up a single thread that is waiting on this object's monitor. If any threads are waiting on this object, one of them is chosen to be awakened. The choice is arbitrary and occurs at the discretion of the implementation. A thread waits on an object's monitor by calling one of the wait methods. The awakened thread will not be able to proceed until the current thread relinquishes the lock on this object. The awakened thread will compete in the usual manner with any other threads that might be actively competing to synchronize on this object; for example, the awakened thread enjoys no reliable privilege or disadvantage in being the next thread to lock this object.

The Sleeping Barber Write a solution to the sleeping barber problem using monitors and condition variables The sleeping barber: The barber cuts peoples hair in his shop, which has 2 doors – entrance and exit. When people are in his shop, he gives them a hair cut, one at a time. When none are in his shop, he sleeps on his chair. When a customer arrives and finds the barber sleeping, he awakens him and sits in the barber’s chair to receive his haircut. After the cut is done, the barber sees the customer out through the exit door. If the barber is busy when a customer arrives, the customer waits in one of the chairs in the shop. If all are occupied, he goes away. After serving a customer the barber looks to see if any are waiting and if so proceeds to serve one of them. Otherwise, he sleeps again in his chair.

The Sleeping Barber barbershop: monitor waiting : integer := 0; % customers waiting for haircut customers : condition; % used by barber, wait for a customer barber : condition; % used by customer, wait for barber procedure seek-customer( ) % called by the barber begin if waiting==0 then WAIT (customers); % sleeps if no customers waiting = waiting-1; % one less customer waiting cut-hair(); SIGNAL (barber); % free a waiting customer end seek-customer;

The Sleeping Barber procedure get-haircut( ) % called by a customer begin % is there a free chair to sit and wait? % if no free chairs just go away if waiting < chairs then { waiting = waiting+1; % one more customer waiting SIGNAL (customers) % if the barber is asleep WAIT (barber); % wait for turn with the barber } end get-haircut; end barbershop; % End of monitor

Producer Consumers with Monitor Monitor ProducerConsumer 1. condition full, empty 2. integer count initially 0 3. procedure insert(item: integer) 4. begin 5. if count=N then wait(full) 6. insert_item(item) 7. count=count+1 8. signal(empty) 9. end 10. procedure remove: integer 11. begin 12. if count=0 then wait(empty) 13. remove=remove_item() 14. count=count-1 15. signal(full) 16. end end Monitor

Producer Consumers with Monitor Monitor ProducerConsumer 1. condition full, empty 2. integer count initially 0 3. procedure insert(item: integer) 4. begin 5. if count=N then wait(full) 6. insert_item(item) 7. count=count+1 8. signal(empty) 9. end 10. procedure remove: integer 11. begin 12. if count=0 then wait(empty) 13. remove=remove_item() 14. count=count-1 15. signal(full) 16. end end Monitor Will it work with Mesa type monitor? What about Hoare?

Java and monitors – Exercise Write a code snippet in Java which will enforce a FIFO waking order (i.e., create a class in Java that will allow a programmer fair synchronization)

Spurious wakeups On some threading API’s (e.g., for linux and windows) monitors and conditional variables are vulnerable to spurious wakeups. Spurious wakeup describes a complication in the use of condition variables in which a thread might be awoken from its waiting state even though no thread signaled the condition variable. Since Java uses the native threads implementation (native for the OS which is running the JVM) one must handle spurious wakeups. For correctness it is necessary, then, to verify that the condition is indeed true after the thread has finished waiting and continue waiting otherwise.

Java and monitors – Solution class SafeMonitor { boolean released = false; // this flag avoids race!!! synchronized void await() throws InterruptedException { while (! released) { wait(); } } synchronized void signal(){ if (! released){ released = true; notify(); } } }

Java and monitors – Solution class CriticalSection { private List<SafeMonitor> waiting; private boolean busy; public CriticalSection() { waiting = new LinkedList<>(); busy = false; } public void enter() { SafeMonitor myLock = null; synchronized (this) { if (! busy) { busy = true; } else { myLock = new SafeMonitor(); waiting.add(myLock); } } if (myLock != null) myLock.await(); public synchronized void leave() { if (!waiting.isEmpty()) { waiting.remove().signal(); } else { busy = false; }

Java and monitors – Solution class CriticalSection { private List<SafeMonitor> waiting; private boolean busy; public CriticalSection() { waiting = new LinkedList<>(); busy = false; } public void enter() { SafeMonitor myLock = null; synchronized (this) { if (! busy) { busy = true; } else { myLock = new SafeMonitor(); waiting.add(myLock); } } if (myLock != null) myLock.await(); public synchronized void leave() { if (!waiting.isEmpty()) { waiting.remove().signal(); } else { busy = false; } Does this code guarantee a FIFO waking order which is equivalent to the order in which threads reached the critical section entrance? What happens when multiple threads attempt to enter at the same time?

The one-way tunnel problem Allows any number of processes in the same direction If there is traffic in the opposite direction – have to wait A special case of readers/writers

The one way tunnel (exam 2004) The one way tunnel solution: int count[2]; Semaphore busy=1, mutex=1; Semaphore waiting[2]={1,1}; void arrive(int direction){ down(&waiting[direction]); down(&mutex) count[direction]+=1; if (count[direction]==1){ up(&mutex) down(&busy); } else up(&waiting[direction]); void leave(int direction){ down(&mutex); count[direction]-=1; if (count[direction]==0){ up(&busy); } up(&mutex);

The one way tunnel (exam 2004) Add changes to the one way tunnel solution so that there will be no starvation. If vehicles are present on both “0” and “1” they will take alternate turns in entering the tunnel. When there are vehicles coming from only one direction, they can pass through with no limitations. Notes: you may only use integers and binary semaphores (can assume fairness).

The one way tunnel (exam 2004) The one way tunnel solution: int count[2]; Semaphore busy=1, mutex=1, new_mutex=1; Semaphore waiting[2]={1,1}; void arrive(int direction){ down(&waiting[direction]); down(&new_mutex); down(&mutex); count[direction]+=1; if (count[direction]==1) { down(&busy); } else up(&mutex) up(&new_mutex); up(&waiting[direction]); void leave(int direction){ down(&mutex); count[direction]-=1; if (count[direction]==0){ up(&busy); } up(&mutex);

One way, convoy (midterm 2008) In the following question you must implement a solution to the convoy problem using only semaphores (and regular variables). In this problem, each thread represent a vehicle. The vehicles must go through a one way tunnel, but unlike the tunnel problem, here vehicles may only cross the tunnel in groups of exactly 5 (all in the same direction). A group of another 5 vehicles (from the same or opposite direction) may cross the tunnel again, only after the previous group of 5 vehicles comes out of it. The general code structure is as follows: Variable Declaration PrepareToCross(int direction) CROSS DoneWithCrossing(int direction)

One way, convoy (midterm 2008) Implement PrepareToCross() and DoneWithCrossing(). For your implementation you may only use semaphores (counting or binary) and regular variables. No busy-waiting is allowed. We say a thread is passing the tunnel if it completed its call to PrepareToCross and haven’t called DoneWithCrossing or if it is still in PrepareToCross but is no longer waiting on a semaphore, and when it will receive CPU time it may complete the procedure without waiting. Your implementation must satisfy the following conditions: Mutual Exclusion – threads from different direction may never be in the tunnel at the same time. Threads may only cross in groups of 5. When the first is leaving PrepareToCross, there are exactly 4 others which are passing the tunnel as well. Progress – Whenever there are 5 (or more) threads from the same direction waiting to cross the tunnel, then eventually, 5 will.

One way, convoy (midterm 2008) We will use the following: Counting Semaphore waitingToCross[]={5,5} Counting Semaphore barrier[]={0,0} Binary Semaphore busy=1 Binary Semaphore mutex=1 int waitingCount[]={0,0} int passed=0

One way, convoy (midterm 2008) PrepareToCross(int i){ down(&waitingToCross[i]); down(&mutex); waitingCount[i]++; If (waitingCount[i]<5){ up(&mutex); down(&barrier[i]); } else { waitingCount[i]=0; down(&busy); for (int k=0; k<4; k++) up(&barrier[i]); } up(&waitingToCross[i]); DoneWithCrossing(int i){ down(&mutex); passed++; if (passed==5){ passed=0; up(&busy); } up(&mutex);

Message passing Used on distributed systems (when there is no shared memory). Uses send(), and receive() system calls. Introduces a new set of problems, such as acknowledgments, sequencing, addressing, authentication, etc’…

Reader/Writer problem with MP Write a solution to the reader/writer problem using Message Passing. Assume the following: Three groups of processes: readers, writer, manager. Multiple readers may access the DB simultaneously. A writer needs exclusive access to the DB. Readers have preference.

Reader/Writer problem with MP Reader: while (true){ SEND (manager, start_read); RECEIVE (manager, msg); % wait for confirmation read_db(); SEND (manager, end_read); use_data(); } Writer: while (true){ generate_data(); SEND (manager, start_write); RECEIVE (manager, msg); % wait for confirmation write_to_db(); SEND (manager, end_write);

Reader/Writer problem with MP Manager: int readers_count=0; % number of readers accessing DB boolean writing=false; % writing flag Message msg; Queue readQ, writeQ; % Queues for waiting readers and writers ProcessID src; % pid while (true){ src = RECEIVE(msg); switch msg.type{ case (start_read): if (not writing){ send(src, ok); readers_count++; } else readQ.add(src);

Reader/Writer problem with MP case (end_read): readers_count--; if (readers_count==0 && not writeQ.empty){ src=writeQ.remove; SEND (src, ok); writing = true; } case (start_write): if (readers_count==0 && not writing){ SEND (src, ok); writing = true; } else writeQ.add(src);

Reader/Writer problem with MP case (end_write): writing = false; if (readQ.empty && not writeQ.empty){ src = writeQ.remove; SEND(src, ok); writing = true; } else { while (not readQ.empty){ src = readQ.remove; send(src, ok); readers_count++; } } } % switch } % while