1. Synchronization II: CPE 261403 - Operating Systems http://www.e-cpe.org/moodle

2. Classical Problems of Synchronization Bounded-Buffer Problem (6.6.1) Dining-Philosophers Problem (6.6.3) (Readers-Writers Problem 6.6.2)

3. Bounded-Buffer Problem while (true) { // produce an item wait (empty); wait (mutex); // add the item to the buffer signal (mutex); signal (full); } while (true) { wait (full); wait (mutex); // remove an item from buffer signal (mutex); signal (empty); // consume the removed item } Producer Consumer 10N MutexFullEmpty Semaphores

4. Human Error is Hard to Debug

5. Monitors Higher level API than Semaphores Can help prevent human error Figure 6.19

6. Bounded Buffer Problem monitor BoundedBuffer { function Produce (value) { if BufferLen < N // Add a value to Buffer } Function consume() { If BufferLen > 0 // remove a value from Buffer } // usage example BoundedBuffer.Produce(x) BoundedBuffer.Consume()

7. Bounded Buffer Problem monitor BoundedBuffer { Condition content Condition emptySpace function Produce (value) { if BufferLen < N // Add a value to Buffer content.signal() else emptySpace.wait() } Function consume() { If BufferLen > 0 // remove a value from Buffer emptySpace.signal() else content.wait() }

8. Counting Semaphores vs Conditions Semaphore i = 0 Signal (i) // i will = 1 Condition i i.signal() // nothing happens // if not waiting // process Counting Semaphores Condition (Less bugs)

9. Dining-Philosophers Problem Figure 6.16

10. A solution using Semaphores Every philosopher runs this code While (true) { wait ( chopstick[i] ); wait ( chopStick[ (i + 1) % 5] ); // eat signal ( chopstick[i] ); signal (chopstick[ (i + 1) % 5] ); // think }

11. Deadlock?

12. Using Monitors monitor DP { enum { THINKING; HUNGRY, EATING) state [5] ; condition self [5]; void pickup (int i) { state[i] = HUNGRY; test(i); if (state[i] != EATING) self [i].wait; } void putdown (int i) { state[i] = THINKING; // test left and right neighbors test((i + 4) % 5); test((i + 1) % 5); }

13. Using Monitors (Cont.) void test (int i) { if ( (state[(i + 4) % 5] != EATING) && (state[i] == HUNGRY) && (state[(i + 1) % 5] != EATING) ) { state[i] = EATING ; self[i].signal () ; } initialization_code() { for (int i = 0; i < 5; i++) state[i] = THINKING; }