Classical Synchronization

Problems

CS 537 - Introduction to Operating Systems

Classical Problem 1

Producer-Consumer

#include <stdio.h>#include <semaphore.h>

#define SIZE 50#define NUM_THREADS 5

struct object* buffer;sem_t mutex, empty, full;

void nap() {int sleepTime = (int)(5 * random());sleep(sleepTime);

}

void enter(struct object item) {empty.down();mutex.down();count++;buffer[in] = item;in = (in + 1) % SIZE;mutex.up();full.up();

}

struct object remove() {full.down();mutex.down();count--;struct object item = buffer[out];out = (out + 1) % SIZE;mutex.up();empty.up();return item;

}

void producer() {struct object newObj;for(;;) {

nap();newObj = getObject();enter(newObj);

}}

void consumer() {struct object item;for(;;) {

nap();item = remove();useItem(item);

}}

int main(int argc, char** argv) {pthread_t threads[NUM_THREADS];int i;

sem_init(&mutex, 0, 1);sem_init(&empty, 0, SIZE);sem_init(&full, 0, 0);

for(i=0; i<NUM_THREADS; i++) { if(i % 2) pthread_create(&threads[i], NULL, (void*)producer, NULL); else pthread_create(&threads[i], NULL, (void*)consumer, NULL);}consumer();

return 0;}

Classical Problem 2

Readers-Writers

Readers-Writers• Several threads accessing a database

• Some threads read the database (readers)

• Some threads write the database (writers)

• Multiple readers can be accessing database at a time if no writers accessing it

• Only one thread can be accessing database if it is a writer

• 2 major variations– no reader waits if no writer in database

• writers can starve

– all readers wait if a writer is waiting• writer has to wait until current reader (or writer)

is finished with database

• readers may starve

• Many more variations to deal with starvation issues

• Following example is for the first variant– starvation is disregarded

#include <stdio.h>

#include <pthreads.h>

#define NUM_THREADS 5

sem_t mutex, database;

void reader() {

while(true) {

nap();

startRead();

nap();

endRead();

}

}

void writer() {

while(true) {

nap();

startWrite();

nap();

endWrite();

}

}

void nap() {

int sleepTime = (int)(5 * random());

sleep(sleepTime);

}

void startRead() {mutex.down();readerCount++;if(readerCount == 1) { database.down(); }mutex.up();

}

void endRead() {mutex.down();readerCount--;if(readerCount == 0) { database.up(); }mutex.up();

}

void startWrite() { database.down(); }void endWrite() { database.up(); }

int main() {pthread_t threads[NUM_THREADS];int i;

sem_init(&mutex, 0, 1);sem_init(&database, 0, 1);

for(i=0; i<NUM_THREADS; i++) { if(i % 2) pthread_create(&threads[i], NULL, (void*)reader, NULL); else pthread_create(&threads[i], NULL, (void*)writer, NULL);}writer();

return 0;}