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Semaphores

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Mohd Arif
Mohd Arif

1) A semaphore consists of a counter, a waiting list, and wait() and signal() methods. Wait() decrements the counter and blocks if it becomes negative, while signal() increments the counter and resumes a blocked process if the counter becomes positive. 2) The dining philosophers problem is solved using semaphores to lock access to shared chopsticks, with one philosopher designated as a "weirdo" to avoid deadlock by acquiring locks in a different order. 3) The producer-consumer problem uses three semaphores - one to limit buffer size, one for empty slots, and one for locks - to coordinate producers adding to a bounded buffer

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Semaphores qA semaphore is an object that consists of a counter, a waiting list of processes and two methods (e.g., functions): signal and wait. semaphore method signal counter method wait waiting list 1
Semaphore Method: wait void wait(sem S) { S.count--; if (S.count < 0) { add the caller to the waiting list; block(); } } qAfter decreasing the counter by 1, if the counter value becomes negative, then vadd the caller to the waiting list, and then vblock itself. 2
Semaphore Method: signal void signal(sem S) { S.count++; if (S.count <= 0) { remove a process P from the waiting list; resume(P); } } qAfter increasing the counter by 1, if the new counter value is not positive, then vremove a process P from the waiting list, vresume the execution of process P, and return 3
Important Note: 1/4 S.count--; S.count++; if (S.count<0) { if (S.count<=0) { add to list; remove P; block(); resume(P); } } qIf S.count < 0, abs(S.count) is the number of waiting processes. qThis is because processes are added to (resp., removed from) the waiting list only if the counter value is < 0 (resp., <= 0). 4
Important Note: 2/4 S.count--; S.count++; if (S.count<0) { if (S.count<=0) { add to list; remove P; block(); resume(P); } } qThe waiting list can be implemented with a queue if FIFO order is desired. qHowever, the correctness of a program should not depend on a particular implementation of the waiting list. qYour program should not make any assumption about the ordering of the waiting list. 5
Important Note: 3/4 S.count--; S.count++; if (S.count<0) { if (S.count<=0) { add to list; remove P; block(); resume(P); } } qThe caller may be blocked in the call to wait(). qThe caller never blocks in the call to signal(). If S.count > 0, signal() returns and the caller continues. Otherwise, a waiting process is released and the caller continues. In this case, two processes continue. 6
The Most Important Note: 4/4 S.count--; S.count++; if (S.count<0) { if (S.count<=0) { add to list; remove P; block(); resume(P); } } qwait() and signal() must be executed atomically (i.e., as one uninterruptible unit). qOtherwise, race conditions may occur. qHomework: use execution sequences to show race conditions if wait() and/or signal() is not executed atomically. 7
Three Typical Uses of Semaphores qThere are three typical uses of semaphores: vmutual exclusion: Mutex (i.e., Mutual Exclusion) locks vcount-down lock: Keep in mind that semaphores have a counter. vnotification: Indicate an event has occurred. 8
Use 1: Mutual Exclusion (Lock) initialization is important semaphore S = 1; int count = 0; Process 1 Process 2 while (1) { while (1) { // do something entry // do something S.wait(); S.wait(); count++; critical sections count--; S.signal(); S.signal(); // do something exit // do something } } qWhat if the initial value of S is zero? qS is a binary semaphore (similar to a lock). 9
Use 2: Count-Down Counter semaphore S = 3; Process 1 Process 2 while (1) { while (1) { // do something // do something S.wait(); S.wait(); at most 3 processes can be here!!! S.signal(); S.signal(); // do something // do something } } qAfter three processes pass through wait(), this section is locked until a process calls signal(). 10
Use 3: Notification semaphore S1 = 1, S2 = 0; process 1 process 2 while (1) { while (1) { // do something // do something S1.wait(); notify S2.wait(); cout << “1”; cout << “2”; S2.signal(); notify S1.signal(); // do something // do something } } qProcess 1 uses S2.signal() to notify process 2, indicating “I am done. Please go ahead.” qThe output is 1 2 1 2 1 2 …… qWhat if both S1 and S2 are both 0’s or both 1’s? qWhat if S1 = 0 and S2 = 1? 11
Lock Example: Dining Philosophers § Five philosophers are in a thinking - eating cycle. § When a philosopher gets hungry, he sits down, picks up two nearest chopsticks, and eats. § A philosopher can eat only if he has both chopsticks. § After eating, he puts down both chopsticks and thinks. § This cycle continues. 12
Dining Philosopher: Ideas qChopsticks are shared outer critical section items (by two philosophers) left chop locked and must be protected. Semaphore C[5] = 1; qEach chopstick has a C[i].wait(); semaphore with initial C[(i+1)%5].wait(); value 1. has 2 chops and eats qA philosopher calls wait() before picks up a C[(i+1)%5].signal(); C[i].signal(); chopstick and calls signal() to release it. inner critical section right chop locked 13
Dining Philosophers: Code semaphore C[5] = 1; philosopher i wait for my left chop while (1) { // thinking C[i].wait(); wait for my right chop C[(i+1)%5].wait(); // eating release my right chop C[(i+1)%5].signal(); C[i].signal(); release my left chop // finishes eating } Does this solution work? 14
Dining Philosophers: Deadlock! § If all five philosophers sit down and pick up their left chopsticks at the same time, this program has a circular waiting and deadlocks. § An easy way to remove this deadlock is to introduce a weirdo who picks up his right chopstick first! 15
Dining Philosophers: A Better Idea semaphore C[5] = 1; philosopher i (0, 1, 2, 3) Philosopher 4: the weirdo while (1) { while (1) { // thinking // thinking C[i].wait(); C[(i+1)%5].wait(); C[(i+1)%5].wait(); C[i].wait(); // eating // eating C[(i+1)%5].signal(); C[i].signal(); C[i].signal(); C[(i+1)%5].signal(); // finishes eating; // finishes eating } } lock left chop lock right chop 16
Dining Philosophers: Questions qThe following are some important questions for you to work on. vWe choose philosopher 4 to be the weirdo. Does this choice matter? vShow that this solution does not cause circular waiting. vShow that this solution will not have circular waiting if we have more than 1 and less than 5 weirdoes. qThese questions may appear as exam problems. 17
Count-Down Lock Example qThe naïve solution to the dining philosophers causes circular waiting. qIf only four philosophers are allowed to sit down, no deadlock can occur. qWhy? If all four of them sit down at the same time, the right-most philosopher can have both chopsticks! qHow about fewer than four? This is obvious. 18
Count-Down Lock Example semaphore C[5]= 1; semaphore Chair = 4; get a chair while (1) { this is a count-down lock // thinking that only allows 4 to go! Chair.wait(); C[i].wait(); C[(i+1)%5].wait(); // eating this is our old friend C[(i+1)%5].signal(); C[i].signal(); Chair.signal(); } release my chair 19
The Producer/Consumer Problem qSuppose we have a circular buffer of n slots. qPointers in (resp., out) points to the first empty (resp., filled) slot. qProducer processes keep adding info into the buffer qConsumer processes keep bounded-buffer retrieving info from the buffer. 20
Problem Analysis qA producer deposits info into Buf[in] and a consumer retrieves info from Buf[out]. qin and out must be advanced. qin is shared among producers. qout is shared among consumers. qIf Buf is full, producers should buffer is implemented be blocked. with an array Buf[ ] qIf Buf is empty, consumers should be blocked. 21
qWe need a sem. to protect the buffer. qA second sem. to block producers if the buffer is full. qA third sem. to block consumers if the buffer is empty. 22
Solution no. of slots semaphore NotFull=n, NotEmpty=0, Mutex=1; producer consumer while (1) { while (1) { NotFull.wait(); NotEmpty.wait(); Mutex.wait(); Mutex.wait(); Buf[in] = x; x = Buf[out]; in = (in+1)%n; out = (out+1)%n; Mutex.signal(); Mutex.signal(); NotEmpty.signal(); NotFull.signal(); } } notifications critical section 23
Question qWhat if the producer code is modified as follows? qAnswer: a deadlock may occur. Why? while (1) { Mutex.wait(); NotFull.wait(); Buf[in] = x; order changed in = (in+1)%n; NotEmpty.signal(); Mutex.signal(); } 24
The Readers/Writers Problem qTwo groups of processes, readers and writers, are accessing a shared resource by the following rules: vReaders can read simultaneously. vOnly one writer can write at any time. vWhen a writer is writing, no reader can read. vIf there is any reader reading, all incoming writers must wait. Thus, readers have higher priority. 25
Problem Analysis qWe need a semaphore to block readers if a writer is writing. qWhen a writer arrives, it must be able to know if there are readers reading. So, a reader count is required which must be protected by a lock. qThis reader-priority version has a problem: bounded waiting condition may be violated if readers keep coming, causing the waiting writers no chance to write. 26
qWhen a reader comes in, it increase the count. qIf it is the 1st reader, waits until no writer is writing, qReads data. qDecreases the counter. qNotifies the writer that no reader is reading if it is the last. 27
qWhen a writer comes in, it waits until no reader is reading and no writer is writing. qThen, it writes data. qFinally, notifies readers and writers that no writer is in. 28
Solution semaphore Mutex = 1, WrtMutex = 1; int RdrCount; reader writer while (1) { while (1) { Mutex.wait(); RdrCount++; if (RdrCount == 1) blocks both readers and writers WrtMutex.wait(); WrtMutex.wait(); Mutex.signal(); // read data // write data Mutex.wait(); RdrCount--; if (RdrCount == 0) WrtMutex.signal(); WrtMutex.signal(); Mutex.signal(); } } 29

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Semaphores

  • 1. Semaphores qA semaphore is an object that consists of a counter, a waiting list of processes and two methods (e.g., functions): signal and wait. semaphore method signal counter method wait waiting list 1
  • 2. Semaphore Method: wait void wait(sem S) { S.count--; if (S.count < 0) { add the caller to the waiting list; block(); } } qAfter decreasing the counter by 1, if the counter value becomes negative, then vadd the caller to the waiting list, and then vblock itself. 2
  • 3. Semaphore Method: signal void signal(sem S) { S.count++; if (S.count <= 0) { remove a process P from the waiting list; resume(P); } } qAfter increasing the counter by 1, if the new counter value is not positive, then vremove a process P from the waiting list, vresume the execution of process P, and return 3
  • 4. Important Note: 1/4 S.count--; S.count++; if (S.count<0) { if (S.count<=0) { add to list; remove P; block(); resume(P); } } qIf S.count < 0, abs(S.count) is the number of waiting processes. qThis is because processes are added to (resp., removed from) the waiting list only if the counter value is < 0 (resp., <= 0). 4
  • 5. Important Note: 2/4 S.count--; S.count++; if (S.count<0) { if (S.count<=0) { add to list; remove P; block(); resume(P); } } qThe waiting list can be implemented with a queue if FIFO order is desired. qHowever, the correctness of a program should not depend on a particular implementation of the waiting list. qYour program should not make any assumption about the ordering of the waiting list. 5
  • 6. Important Note: 3/4 S.count--; S.count++; if (S.count<0) { if (S.count<=0) { add to list; remove P; block(); resume(P); } } qThe caller may be blocked in the call to wait(). qThe caller never blocks in the call to signal(). If S.count > 0, signal() returns and the caller continues. Otherwise, a waiting process is released and the caller continues. In this case, two processes continue. 6
  • 7. The Most Important Note: 4/4 S.count--; S.count++; if (S.count<0) { if (S.count<=0) { add to list; remove P; block(); resume(P); } } qwait() and signal() must be executed atomically (i.e., as one uninterruptible unit). qOtherwise, race conditions may occur. qHomework: use execution sequences to show race conditions if wait() and/or signal() is not executed atomically. 7
  • 8. Three Typical Uses of Semaphores qThere are three typical uses of semaphores: vmutual exclusion: Mutex (i.e., Mutual Exclusion) locks vcount-down lock: Keep in mind that semaphores have a counter. vnotification: Indicate an event has occurred. 8
  • 9. Use 1: Mutual Exclusion (Lock) initialization is important semaphore S = 1; int count = 0; Process 1 Process 2 while (1) { while (1) { // do something entry // do something S.wait(); S.wait(); count++; critical sections count--; S.signal(); S.signal(); // do something exit // do something } } qWhat if the initial value of S is zero? qS is a binary semaphore (similar to a lock). 9
  • 10. Use 2: Count-Down Counter semaphore S = 3; Process 1 Process 2 while (1) { while (1) { // do something // do something S.wait(); S.wait(); at most 3 processes can be here!!! S.signal(); S.signal(); // do something // do something } } qAfter three processes pass through wait(), this section is locked until a process calls signal(). 10
  • 11. Use 3: Notification semaphore S1 = 1, S2 = 0; process 1 process 2 while (1) { while (1) { // do something // do something S1.wait(); notify S2.wait(); cout << “1”; cout << “2”; S2.signal(); notify S1.signal(); // do something // do something } } qProcess 1 uses S2.signal() to notify process 2, indicating “I am done. Please go ahead.” qThe output is 1 2 1 2 1 2 …… qWhat if both S1 and S2 are both 0’s or both 1’s? qWhat if S1 = 0 and S2 = 1? 11
  • 12. Lock Example: Dining Philosophers § Five philosophers are in a thinking - eating cycle. § When a philosopher gets hungry, he sits down, picks up two nearest chopsticks, and eats. § A philosopher can eat only if he has both chopsticks. § After eating, he puts down both chopsticks and thinks. § This cycle continues. 12
  • 13. Dining Philosopher: Ideas qChopsticks are shared outer critical section items (by two philosophers) left chop locked and must be protected. Semaphore C[5] = 1; qEach chopstick has a C[i].wait(); semaphore with initial C[(i+1)%5].wait(); value 1. has 2 chops and eats qA philosopher calls wait() before picks up a C[(i+1)%5].signal(); C[i].signal(); chopstick and calls signal() to release it. inner critical section right chop locked 13
  • 14. Dining Philosophers: Code semaphore C[5] = 1; philosopher i wait for my left chop while (1) { // thinking C[i].wait(); wait for my right chop C[(i+1)%5].wait(); // eating release my right chop C[(i+1)%5].signal(); C[i].signal(); release my left chop // finishes eating } Does this solution work? 14
  • 15. Dining Philosophers: Deadlock! § If all five philosophers sit down and pick up their left chopsticks at the same time, this program has a circular waiting and deadlocks. § An easy way to remove this deadlock is to introduce a weirdo who picks up his right chopstick first! 15
  • 16. Dining Philosophers: A Better Idea semaphore C[5] = 1; philosopher i (0, 1, 2, 3) Philosopher 4: the weirdo while (1) { while (1) { // thinking // thinking C[i].wait(); C[(i+1)%5].wait(); C[(i+1)%5].wait(); C[i].wait(); // eating // eating C[(i+1)%5].signal(); C[i].signal(); C[i].signal(); C[(i+1)%5].signal(); // finishes eating; // finishes eating } } lock left chop lock right chop 16
  • 17. Dining Philosophers: Questions qThe following are some important questions for you to work on. vWe choose philosopher 4 to be the weirdo. Does this choice matter? vShow that this solution does not cause circular waiting. vShow that this solution will not have circular waiting if we have more than 1 and less than 5 weirdoes. qThese questions may appear as exam problems. 17
  • 18. Count-Down Lock Example qThe naïve solution to the dining philosophers causes circular waiting. qIf only four philosophers are allowed to sit down, no deadlock can occur. qWhy? If all four of them sit down at the same time, the right-most philosopher can have both chopsticks! qHow about fewer than four? This is obvious. 18
  • 19. Count-Down Lock Example semaphore C[5]= 1; semaphore Chair = 4; get a chair while (1) { this is a count-down lock // thinking that only allows 4 to go! Chair.wait(); C[i].wait(); C[(i+1)%5].wait(); // eating this is our old friend C[(i+1)%5].signal(); C[i].signal(); Chair.signal(); } release my chair 19
  • 20. The Producer/Consumer Problem qSuppose we have a circular buffer of n slots. qPointers in (resp., out) points to the first empty (resp., filled) slot. qProducer processes keep adding info into the buffer qConsumer processes keep bounded-buffer retrieving info from the buffer. 20
  • 21. Problem Analysis qA producer deposits info into Buf[in] and a consumer retrieves info from Buf[out]. qin and out must be advanced. qin is shared among producers. qout is shared among consumers. qIf Buf is full, producers should buffer is implemented be blocked. with an array Buf[ ] qIf Buf is empty, consumers should be blocked. 21
  • 22. qWe need a sem. to protect the buffer. qA second sem. to block producers if the buffer is full. qA third sem. to block consumers if the buffer is empty. 22
  • 23. Solution no. of slots semaphore NotFull=n, NotEmpty=0, Mutex=1; producer consumer while (1) { while (1) { NotFull.wait(); NotEmpty.wait(); Mutex.wait(); Mutex.wait(); Buf[in] = x; x = Buf[out]; in = (in+1)%n; out = (out+1)%n; Mutex.signal(); Mutex.signal(); NotEmpty.signal(); NotFull.signal(); } } notifications critical section 23
  • 24. Question qWhat if the producer code is modified as follows? qAnswer: a deadlock may occur. Why? while (1) { Mutex.wait(); NotFull.wait(); Buf[in] = x; order changed in = (in+1)%n; NotEmpty.signal(); Mutex.signal(); } 24
  • 25. The Readers/Writers Problem qTwo groups of processes, readers and writers, are accessing a shared resource by the following rules: vReaders can read simultaneously. vOnly one writer can write at any time. vWhen a writer is writing, no reader can read. vIf there is any reader reading, all incoming writers must wait. Thus, readers have higher priority. 25
  • 26. Problem Analysis qWe need a semaphore to block readers if a writer is writing. qWhen a writer arrives, it must be able to know if there are readers reading. So, a reader count is required which must be protected by a lock. qThis reader-priority version has a problem: bounded waiting condition may be violated if readers keep coming, causing the waiting writers no chance to write. 26
  • 27. qWhen a reader comes in, it increase the count. qIf it is the 1st reader, waits until no writer is writing, qReads data. qDecreases the counter. qNotifies the writer that no reader is reading if it is the last. 27
  • 28. qWhen a writer comes in, it waits until no reader is reading and no writer is writing. qThen, it writes data. qFinally, notifies readers and writers that no writer is in. 28
  • 29. Solution semaphore Mutex = 1, WrtMutex = 1; int RdrCount; reader writer while (1) { while (1) { Mutex.wait(); RdrCount++; if (RdrCount == 1) blocks both readers and writers WrtMutex.wait(); WrtMutex.wait(); Mutex.signal(); // read data // write data Mutex.wait(); RdrCount--; if (RdrCount == 0) WrtMutex.signal(); WrtMutex.signal(); Mutex.signal(); } } 29