1. Chapter 6 Synchronization Dr. Yingwu Zhu

2. The Problem with Concurrent Execution Concurrent processes (& threads) often access shared data and resources – Need controlled access to the shared data; otherwise result in an inconsistent view of this data Maintaining data consistency must ensure orderly execution of cooperating processes

3. Simple Example Threads T1 and T2 are executing the same procedure using the same variables in and out Threads can be interrupted anywhere What can happen? Shared variables in, out; Procedure Echo(); read (in, keyboard); out := in; write (out, screen); end Echo;

4. Simple Example Threads T1 and T2 are executing the same procedure using the same variables in and out Threads can be interrupted anywhere What can happen? Shared variables in, out; Procedure Echo(); read (in, keyboard); out := in; write (out, screen); end Echo;

5. Race Condition & Critical-Section Problem Race condition – Multiple processes manipulate same data concurrently – The outcome of execution depends on the particular order in which the data access takes place Critical Section – Multiple processes has a segment of code, called critical section which handle race condition – Admission to the critical section MUST be coordinated

6. Race Condition: Example count++ could be implemented as register1 = count register1 = register1 + 1 count = register1 count-- could be implemented as register2 = count register2 = register2 - 1 count = register2 Consider this execution interleaving with “count = 5” initially: S0: producer execute register1 = count {register1 = 5} S1: producer execute register1 = register1 + 1 {register1 = 6} S2: consumer execute register2 = count {register2 = 5} S3: consumer execute register2 = register2 - 1 {register2 = 4} S4: producer execute count = register1 {count = 6 } S5: consumer execute count = register2 {count = 4}

7. Critical-Section Problem Each process looks like: do { CRITICAL SECTION REMAINDER SECTION } while (TRUE); Entry Section Exit Section

8. Solution to Critical-Section Problem MUST satisfy the following three requirements: 1.Mutual Exclusion - If process P i is executing in its critical section, then no other processes can be executing in their critical sections 2.Progress - If no process is executing in its critical section and there exist some processes that wish to enter their critical section, then the selection of the processes that will enter the critical section next cannot be postponed indefinitely 3.Bounded Waiting - A bound must exist on the number of times that other processes are allowed to enter their critical sections after a process has made a request to enter its critical section and before that request is granted Assume that each process executes at a nonzero speed No assumption concerning relative speed of the N processes

9. Types of Solutions Software solutions – Algorithms whose correctness does not depend on any assumptions other than positive processing speed (NO failures!) – Busy waiting Hardware solutions – Rely on some special machine instructions OS solutions – Extending hardware solutions to provide some functions and data structure support to the programmer

10. Software Solution (incomplete) Only 2 processes Local process is i, the other is j What is good? What is bad? Shared flag: array[0..1] of boolean = [false, false]; Repeat flag[i] := true; while flag[j] do no-op; critical section; flag[i] := false; remainder section; Forever

11. Software Solution (incomplete) Only 2 processes Local process is i, the other is j Mutual exclusion – good Progress – bad, if both executed the 2 nd line Bounded time – irrelvant Shared flag: array[0..1] of boolean = [false, false]; Repeat flag[i] := true; while flag[j] do no-op; critical section; flag[i] := false; remainder section; Forever

12. Software Solution (Peterson’s) Only 2 processes Local process is i, the other is j Mutual exclusion – good Progress – good Bounded time – good Shared turn: (0..1) = 0; Shared flag: array[0..1] of boolean = [false, false]; Repeat flag[i] := true; turn := j; while (flag[j] && turn = j) do no-op; critical section; flag[i] := false; remainder section; Forever

13. Peterson’s Solution A classic software-based solution to the critical-section problem Two process solution Assume that the LOAD and STORE instructions are atomic; that is, cannot be interrupted. The two processes share two variables: – int turn; – Boolean flag[2] The variable turn indicates whose turn it is to enter the critical section. If turn == i, then process P i is allowed to execute in its critical section The flag array is used to indicate if a process is ready to enter the critical section. flag[i] = true implies that process P i is ready!

14. Hardware Solution – Disable Interrupts Correct solution for uni- processor machine – Atomic instructions During critical section, multiprogramming is not utilized  perf. penalty Too inefficient for multi- processor machines, interrupt disabling message passing to all processors, delay entrance to critical section Repeat disable interrupts; critical section; enable interrupts; remainder section; Forever

15. Hardware Solution – Test and Set Test and modify the content of a word atomically boolean TestAndSet (boolean *target) { boolean rv = *target; *target = TRUE; return rv: } Shared boolean lock:= false; Repeat while ( TestAndSet (&lock )); critical section lock = FALSE; remainder section Forever

16. OS Solution - Semaphores Synchronization tool that does not require busy waiting 2 atomic operations that logically look like this: S: integer wait (S) { while S <= 0; // no-op S--; } signal (S) { S++; } semaphore S; // initialized to 1 wait (S); Critical Section signal (S);

17. Semaphore Implementation With each semaphore there is an associated waiting queue. Each semaphore has two data items: typedef struct { int value; struct PCB *list; } semaphore; Two operations: – Block : place the process invoking the operation on the waiting queue. – Wakeup: remove one of processes in the waiting queue and place it in the ready queue.

18. Semaphore Implementation Implementation of wait: wait (S){ value--; if (value < 0) { add this process P to waiting queue block(P); } Implementation of signal: Signal (S){ value++; if (value <= 0) { remove a process P from the waiting queue wakeup(P); }

19. Semaphore Properties S.value >= 0: value is the # of additional process that can execute wait(s) w/o blocking S.value < 0: abs(value) is the # of processes blocked on the waiting queue Atomicity: – No two processes in wait() and signal() at the same time – No even one in wait() and one in signal(), even with multiple CPUs – Uniprocessor: disable interrupts – Multiprocessor: software or hardware scheme

20. Semaphore Types Counting semaphore – integer value can range over an unrestricted domain – Used to control access to a given resource consisting of a finite number of instances – The semaphore is initialized to the number of resources available Binary semaphore – integer value can range only between 0 and 1; can be simpler to implement – Also known as mutex locks

21. Producer-Consumer Problem Also called the bounded-buffer problem A producer produces data that is consumed by a consumer A buffer holds data that holds produced data that is not yet consumed! Multiple producers and consumers Initially we assume the buffer is unbounde d 0 1 234 out in ……

22. Producer-Consumer: unbounded buffer void append(int d) { buffer[in++] = d; } int take() { int x = buffer[out]; out++; return x; } Initialization: semaphores: mutex = 1, full = 0; integers: int = 0, out = 0; Producer: While (1) { produce x; wait(mutex); append(x); signal(mutex); signal(full); } Consumer: While (1) { wait(full); wait(mutex); x = take(); signal(mutex); consume x; } [1] Does it matter to switch two signal() in producer? [2] Does it matter to switch two signal() in consumer?

23. Producer-Consumer: Bounded Buffer As before, the consumer can take when the number of data items is >=1 This time, the producer can append when the number of empty slots >=1 Can you modify the previous code to accommodate a bounded buffer? Hint: the consumer produces empty slots! Cyclic buffer: 0 1 234 out in

24. Producer-Consumer: bounded buffer void append(int d) { buffer[in] = d; in = (in + 1) % N; } int take() { int x = out; out = (ouit+1) %N; return buffer[x]; } Initialization: semaphores: mutex = 1, full = 0; empty = N; integers: int = 0, out = 0; Producer: While (1) { produce x; wait(empty); wait(mutex); append(x); signal(mutex); signal(full); } Consumer: While (1) { wait(full); wait(mutex); x = take(); signal(mutex); signal(empty); consume x; }

25. Readers-Writers Problem A data set is shared among a number of concurrent processes – Readers – only read the data set; they do not perform any updates – Writers – can both read and write. Problem – allow multiple readers to read at the same time. Only one single writer can access the shared data at the same time (if on writer access it, no readers or other writers are allowed to access the data). Shared Data – Data set Using semaphores – What semaphores do we need?

26. Readers-Writers Problem – Semaphore mutex initialized to 1. – Semaphore wrt initialized to 1. – Integer readcount initialized to 0.

27. Readers-Writers Problem (Cont.) Exercise: Implement readers and writers

28. Readers-Writers Problem Initialization: semaphores: mutex = 1, wrt = 1; integers: readcount = 0; Reader: wait(mutex); readcount++; If (readcount==1) wait(wrt); signal(mutex); read; wait(mutex); readcount--; If(readcount==0) signal(wrt); signal(mutex); Writer: wait(wrt); write; signal(wrt); }