ample code for the reader-writer: #include #include #include /* This program provides a possible solution for first readers writers problem using mutex and semaphore. I have used 10 readers and 5 producers to demonstrate the solution. You can always play with these values. / sem_t wrt; pthread_mutex_t mutex; int cnt = 1; int numreader = 0; void writer(void wno) { sem_wait(&wrt); cnt = cnt2; printf("Writer %d modified cnt to %d\n",(((int )wno)),cnt); sem_post(&wrt); } void reader(void rno) { // Reader acquire the lock before modifying numreader pthread_mutex_lock(&mutex); numreader++; if(numreader == 1) { sem_wait(&wrt); // If this id the first reader, then it will block the writer } pthread_mutex_unlock(&mutex); // Reading Section printf("Reader %d: read cnt as %d\n",((int )rno),cnt); // Reader acquire the lock before modifying numreader pthread_mutex_lock(&mutex); numreader--; if(numreader == 0) { sem_post(&wrt); // If this is the last reader, it will wake up the writer. } pthread_mutex_unlock(&mutex); } int main() { pthread_t read[10],write[5]; pthread_mutex_init(&mutex, NULL); sem_init(&wrt,0,1); int a[10] = {1,2,3,4,5,6,7,8,9,10}; //Just used for numbering the producer and consumer for(int i = 0; i < 10; i++) { pthread_create(&read[i], NULL, (void )reader, (void )&a[i]); } for(int i = 0; i < 5; i++) { pthread_create(&write[i], NULL, (void )writer, (void )&a[i]); } for(int i = 0; i < 10; i++) { pthread_join(read[i], NULL); } for(int i = 0; i < 5; i++) { pthread_join(write[i], NULL); } pthread_mutex_destroy(&mutex); sem_destroy(&wrt); return 0; } Consider the sample code for the reader-writer problem below. Change this code by adding counting semaphores to implement the scenario where the number of active readers is upper-bounded by N.

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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ample code for the reader-writer:

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

/*
This program provides a possible solution for first readers writers problem using mutex and semaphore.
I have used 10 readers and 5 producers to demonstrate the solution. You can always play with these values.
*/

sem_t wrt;
pthread_mutex_t mutex;
int cnt = 1;
int numreader = 0;

void *writer(void *wno)
{
sem_wait(&wrt);
cnt = cnt*2;
printf("Writer %d modified cnt to %d\n",(*((int *)wno)),cnt);
sem_post(&wrt);

}
void *reader(void *rno)
{
// Reader acquire the lock before modifying numreader
pthread_mutex_lock(&mutex);
numreader++;
if(numreader == 1) {
sem_wait(&wrt); // If this id the first reader, then it will block the writer
}
pthread_mutex_unlock(&mutex);
// Reading Section
printf("Reader %d: read cnt as %d\n",*((int *)rno),cnt);

// Reader acquire the lock before modifying numreader
pthread_mutex_lock(&mutex);
numreader--;
if(numreader == 0) {
sem_post(&wrt); // If this is the last reader, it will wake up the writer.
}
pthread_mutex_unlock(&mutex);
}

int main()
{

pthread_t read[10],write[5];
pthread_mutex_init(&mutex, NULL);
sem_init(&wrt,0,1);

int a[10] = {1,2,3,4,5,6,7,8,9,10}; //Just used for numbering the producer and consumer

for(int i = 0; i < 10; i++) {
pthread_create(&read[i], NULL, (void *)reader, (void *)&a[i]);
}
for(int i = 0; i < 5; i++) {
pthread_create(&write[i], NULL, (void *)writer, (void *)&a[i]);
}

for(int i = 0; i < 10; i++) {
pthread_join(read[i], NULL);
}
for(int i = 0; i < 5; i++) {
pthread_join(write[i], NULL);
}

pthread_mutex_destroy(&mutex);
sem_destroy(&wrt);

return 0;

}

Consider the sample code for the reader-writer problem below. Change this code by adding counting semaphores to implement the scenario where the number of active readers is upper-bounded by N.

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