Class Announcement TA is expected to add/move office hours to tomorrow for Project 0 Project 0 deadline is extended to next Monday April 13 noon. A copy.

Class Announcement

• TA is expected to add/move office hours to tomorrow for Project 0

• Project 0 deadline is extended to next Monday April 13 noon.

• A copy of the text book reserved in the library.

• Any Issues for Project 0? Pass all public tests locally and then fail in

submit.cs?

Testfile2 without/with an invalid command

ls -l > tempwc templs -l -f > tempwc < temp > temp2sort < temp > temp3cat temp2diff temp2 temp3 > temp4wc temp4exit

ls -l > tempwc templs -l -f > tempwc < temp > temp2sort < temp > temp3catt temp2diff temp2 temp3 > temp4wc temp4exit

Testfile4 without/with an invalid command

ls -l -f | sort > templs -l | sort > temp1diff temp temp1 > temp2cat < temp | uniq | sort > temp3diff temp3 temp |sort | wc > temp4cat temp4exit

ls -l -f | sort > templs -l | sort > temp1diff temp temp1 > temp2cat < temp | uniq | sort > temp3catt temp3 | wcdiff temp3 temp |sort | wc > temp4cat temp4exit

Autograding only compares with expected answers for processing all valid commands. Donot test weird cases.

Lecture 5: Process/Thread Synchronization

CS170. UCSB.2015. T. Yang

Topics

• The critical-section problem and examples

• Solutions Locks Semaphores Condition variables (monitors).

• Synchronization in Nachos and Pthreads

Application example that needs synchronization2 producer threads and 2 consumer threads

Shared data

Shared variables in a producer/consumer program

count -- number of items available in -- points to the writeable slot out -- points to the slot to read

in out

Producer Thread Code

while (true) {

while (count == BUFFER_SIZE) ;

// buffer is full, spin

buffer [in] = nextProduced;

in = (in + 1) % BUFFER_SIZE;

count++;

}

in

Consumer thread

while (true) {

while (count == 0) ; // Nothing is available. Spin

nextConsumed = buffer[out];

out = (out + 1) % BUFFER_SIZE;

count--;

}

out

Race Conditions: Example

Count--;

Producer thread Consumer thread

Count++;

Count=5

Is count still 5?

Race Conditions: two or more processes (threads) are reading and writing on shared data and the final result depends on who runs precisely & when

Review compiled code

Count—: register2 = count register2 = register2 - 1 count = register2


Count++: register1 = count register1 = register1 + 1 count = register1

Count=5

Is count still 5?

Executing compiled code

Count—:

register2 = count register2 = register2 - 1 count = register2


Count++: register1 = count register1 = register1 + 1 count=register1

Count=5

Time Is count still 5? 4, 6?Is count still 5? 4, 6?

Executing compiled code

Count—: register2 = count

register2 = register2 - 1

count = register2


Count++: register1 = count

register1 = register1 + 1

count = register1

Count=5

Is count still 5? 4, 6?Is count still 5? 4, 6?Time

Critical-Section Management

Critical section: the part of the program where shared variables are accessed

Property of Critical-Section Solutions

1.Mutual Exclusion – Only one can enter the critical section. Safety property in AD book

2.Progress - If some processes wish to enter their critical section and nobody is in the critical section, then one of them will enter in a limited time. Liveness property in AD book

3.Bounded Waiting - If one process starts to wait for entering an critical section, there is a limit on the number of times other processes entering the section before this process enters. Liveness property in AD book

1.Locks

2.Semaphore

3.Conditional Variables

Solution to Critical-section Problem

Acquire lock

Critical section

Release lock

Lock for Critical-section Problem

•Thread waits if there is a lock.•It enters the critical after acquiring a lock.•Only the thread who locks can unlock.

Nachos Lock Interface

• Lock can be in one of two states: locked or unlocked

• class Lock {

public:

Lock (char *name);

//create a lock with unlocked state

~Lock();

void Acquire(); //Atomically waits until the lock is unlocked, and then sets the lock to be locked.

void Release(); //Atomically changes the state to be unlocked. Only the thread who owns the lock can release.

}

How to use locks

• Typically associate a lock with a piece of shared data for mutual execlusion.

• When a thread needs to access, it first acquires the lock, and then accesses data. Once access completes, it releases the lock.

Example

Lock *k= new Lock (“Lock”);

k->Acquire();

Perform critical section operations.

k->Release();

}

Synchronization with Locks: Example

int a=0;Lock *k;void sum (int p){ int c; k->Acquire();

a=a+1;c=a;k->Release()printf(“%d : a= %d\n”, p,c);

}

void main() {

Thread *t=new Thread (“Child”);

k=new Lock(“Lock”);

t->Fork(sum,1);

sum(0);

}

Semaphore for the critical-section problem

• Semaphore S – nonnegative integer variable• Can only be accessed /modified via two indivisible

(atomic) operations wait (S) { //also called P() while S <= 0

; // wait in a queue S--; } signal (S) { //also called V() //wake up some waiting thread S++; }

Semaphore

• Counting semaphore – initial value representing how many threads can be in the critical section. Binary semaphore (also known as mutex locks) – integer value

ranged between 0 and 1.• Provides mutual exclusion

Semaphore mutex; // initialized to 1

do {

P(mutex);

// Critical Section

V(mutex);

} while (TRUE);• Not recommended by the AD text book (Chapter 5.8), but still

widely used.

Semaphores in Nachos

• class Semaphore{

public:

Semaphore (char *name, int counter);

// initial counter value

~Semaphore();

void P(); //Atomically waits until the counter is greater than 0 and then decreases the counter

void V(); //Atomically increases the counter

}

Synchronization with Semaphore: Example

int a=0;Semaphore *s;void sum (int p){ int c; s->P();

a=a+1;c=a;s->V();printf(“%d : a= %d\n”, p,c);

}

void main() {

Thread *t=new Thread (“Child”);

s=new Semaphore(“S”,1);

t->Fork(sum,1);

sum(0);

}

Synchronization in Consumer thread

while (true) {

while (count == 0) ; // sping waiting



count--;

Use data for other things

}

out

Synchronization in Producer/ Consumer threads

Consumer loop:

while (count == 0) ;



count--;

Use data and do others

Producer loop:

Produce next item




count++;

out

What is shared?How to synchronize?

count

in

Synchronization AttemptConsumer loop:




count--;

Use dataProducer loop:

Produce next item




count++;

out

incount

Deadlock?How to avoid spin?

Another Synchronization AttemptConsumer loop:




count--;


Produce next item




count++;

out

incount

Mutual execution?

in

What can happen with this trace?

Consumer 1 :




count--;

Do other things

out

incount

in

Consumer 2:




count--;

Use data and do others

Semaphore for consumer/producer problem

• Simplify the problem: Unbounded buffer Worry about bounded buffer constraint later

• Condition to check: Is there a data item available to consume? Not available? Wait using 1 semaphore

Semaphore for consumer-producer problem with unbounded buffer

Semaphore *data =new Semaphore(“Data”, 0);– Indicate # of data items available for consuming

Consumer thread: while(1) {

data-> P(); //wait until an item is available

Use data item;}

Producer thread: while(1){

produce next item;data->V(); //notify an item is available

}

Semaphore can do more than mutual exclusion, and can synchronize a consumer/producer pipe

Semaphore for consumer/producer problem with bounded buffer

What condition to check for bounded buffer?

Conditions to check: Is there a data item available to consume? Is there space available to produce a data

item? Wait using two semaphores

Consumer-producer problem with bounded buffer

• Two semaphores: Semaphore *data =new Semaphore(“Data”, 0);

– Indicate # of data items available for consuming Semaphore *space =new Semaphore(“Space”, 10); Indicate # of 10 data slots available.

Producer thread:

while(1){

produce next item;

data->V(); //notify item is available

}


• Two semaphores: Semaphore *data =new Semaphore(“Data”, 0);

– Indicate # of data items available for consuming Semaphore *space =new Semaphore(“Space”, 10); Indicate # of 10 data slots available.

Producer thread:

while(1){

space->P();//wait until space is available

produce next item;

data->V(); //notify data is available

}


Semaphore *data =new Semaphore(“Data”, 0); Semaphore *space =new Semaphore(“Space”, 10); Producer thread:

while(1){


produce next item;


} Consumer thread:

while(1){

data->P();//wait until data is available

consume next item;

}


Semaphore *data =new Semaphore(“Data”, 0); Semaphore *space =new Semaphore(“Space”, 10); Producer thread:

while(1){


produce next item;


} Consumer thread:

while(1){

data->P();//wait until data is available

consume next item;

space->V();//notify space is available

}

Synchronization for pthreads: Mutex

pthread_mutex_t mutex; const pthread_mutexattr_t attr;int status;

status = pthread_mutex_init(&mutex,&attr);

status = pthread_mutex_destroy(&mutex);

status = pthread_mutex_unlock(&mutex);

status = pthread_mutex_lock(&mutex); - block

Thread i

……lock(mutex)……critical region……unlock(mutex)……

Semaphore for Pthreads

int status,pshared;

sem_t sem;

unsigned int initial_value;

status = sem_init(&sem,pshared,initial_value);

status = sem_destroy(&sem);

status = sem_post(&sem);

status = sem_wait(&sem);

What now?Consumer loop:




count--;


Produce next item




count++;

out

incount

How to use semaphore with lock?

Synchronization with semaphoreConsumer loop:

data->P();



count--;

space->V();


Produce next item

space->P();//wait until space is available) ;



count++;


Synchronization with semaphore/lock?Consumer loop:

data->P();



count--;

space->V();


Produce next item




count++;


What happens when a consumer does not hod a lock after waking up from data->p()?

Consumer 1:

data->P();



count--;

space->V();

Use data

Consumer 2:

data->P();



count--;

space->V();

Use data

Synchronization with semaphore/lock?Consumer loop:

data->P();



count--;

space->V();


Produce next item




count++;


How to hold a lock immediately after waking up?

Solution to Critical-section Problem

1.Locks

2.Semaphore

3.Conditional Variables

Condition Variables in Nachos• Pair it with a lock

• class Condition {

public:

Condition(char *name);

~Condition();

void Wait(Lock *mylock); //Atomically releases the lock and waits. When it is returned, the lock is reacquired again.

void Signal(Lock *myLock); //Wake up one waiting thread to run. The lock is not released.

void Broadcast(Lock *myLock); //Wake up all threads waiting on the condition. The lock is not released.

}

Condition Variables for consumer-producer problem with unbounded buffer

• int avail=0; // # of data items available for consumption• Lock *L; Condition *cv= new Condition (“condition”);• Consumer thread loop:

L.Acquire();if (avail <=0) cv.Wait(L);

Fetch next item; avail = avail-1;L.Release();Do other things

Producer thread loop: L.Acquire();Add next item; avail = avail+1; cv.Signal(L); //notify an item is availableL.Release();Do other things

Can conditionstill be true afterwake-up?How to protect?

Condition Variables for consumer-producer problem with unbounded buffer

• int avail=0; // # of data items available for consumption• Lock *L; Condition *cv= new Condition (“condition”);• Consumer thread loop:

L.Acquire();while(avail <=0) cv.Wait(L);

Fetch next item; avail = avail-1;L.Release();Do other things

Producer thread loop: L.Acquire();Add next item; avail = avail+1; cv.Signal(L); //notify an item is availableL.Release();Do other things

Thread 1

mylock.Acquire();

When condition is not satisfied,

cv.Wait(mylock);

Critical Section;

mylock.Release(); Thread 2:

mylock.Acquire();

When condition can satisfy,

cv.Signal(mylock);

mylock.Relase();

How to Use Condition Variables: Typical Flow

When to use condition broadcast()?

• When waking up one thread to run is not sufficient.

• Example: concurrent malloc()/free() for allocation and deallocation of objects with non-uniform sizes.

malloc()/free() with condition broadcast• Initially 10 bytes are free. • m() stands for malloc(). f() for free()

Thread 1:

m(10) – succ

f(10) –broadcast?

m(7) – wait

Resume m(7)-wait

Thread 2:

m(5) – wait

Resume m(5)-succ

f(5) –broadcast?

Thread 3:

m(5) – wait

Resume m(5)-succ

m(3) –wait

Resume m(3)-succ

malloc()/free() with condition signal• Initially 10 bytes are free. • m() stands for malloc(). f() for free()

Thread 1:

m(10) – succ

f(10) –signal?

m(7) – wait

Resume m(7)-wait

Thread 2:

m(5) – wait

Resume m(5)-succ

f(5) –signal?

Thread 3:

m(5) – wait

Resume m(5)-succ

m(3) –wait

Resume m(3)-succ

Pthread synchronization: Condition variables

int status;

pthread_condition_t cond;

const pthread_condattr_t attr;

pthread_mutex mutex;

status = pthread_cond_init(&cond,&attr);

status = pthread_cond_destroy(&cond);

status = pthread_cond_wait(&cond,&mutex);

status = pthread_cond_signal(&cond);

status = pthread_cond_broadcast(&cond);

Summary

• Concurrent threads are a very useful abstraction Allow transparent overlapping of computation and I/O Allow use of parallel processing when available

• Concurrent threads introduce problems when accessing shared data Programs must be insensitive to arbitrary interleaving Without careful design, shared variables can become

completely inconsistent• Synchronization with lock, condition variables, &

semaphores

Class Announcement TA is expected to add/move office hours to tomorrow for Project 0 Project 0 deadline is extended to next Monday April 13 noon. A copy.

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Class Announcement TA is expected to add/move office hours to tomorrow for Project 0 Project 0 deadline is extended to next Monday April 13 noon. A copy.