CS3413/Lab4/Lab4.c

/*
 * Question 2: You see the number 1 most of the time. It is sometimes different because there are periods of time when
 * the threads are reading/writing to the same place, but with the number of increments being 100, it is not likely that
 * the threads will be reading/writing at the same time. In the CPU what happens is that a value is loaded into a
 * register, while another thread completes a task, and then the value is written back to the memory. This leads to
 * missing increments/decrements.
 *
 * Question 4: You see the number 1 often. Since the number of increments is 1000, the chance has increased for threads
 * to be reading/writing at the same time. This leads to more missing increments/decrements.
 *
 * Question 6: You see the number 1 almost never. The number of loops at this point almost guarantees that the threads
 * will be reading/writing at the same time, and the counter missing increments/decrements.
 *
 * Question 8: The number 1 is the only thing displayed. This is because the mutex lock ensures that the threads will
 * not be reading/writing at the same time, and the counter will not miss increments/decrements. When another thread
 * tried to lock the mutex, it waits for it to first become unlocked, ensuring proper thread synchronization.
 *
 * Question 13: The program no longer finishes, because the semaphore is waiting for the other thread to increment it,
 * but the other thread finishes incrementing it before the first thread gets the chance to print more than 3 times.
 * The minus function prints twice at the end because the print comes before the semaphore wait, so there is one minus
 * for each plus, plus one minus before it has to wait for the semaphore. This situation is called a deadlock.
 */


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

#define RANDOM_WITHIN_RANGE(a, b, seed) (rand_r(&seed)%b+a)

int gv = 1;
pthread_mutex_t mutex;
sem_t sem;

#define loop 10000

void *inc() {
	for (int i = 0; i < loop; i++) {
		pthread_mutex_lock(&mutex);
		gv++;
		pthread_mutex_unlock(&mutex);
	}
	return NULL;
}

void *dec() {
	for (int i = 0; i < loop; i++) {
		pthread_mutex_lock(&mutex);
		gv--;
		pthread_mutex_unlock(&mutex);
	}
	return NULL;
}

void *minus() {
	int i = 0;
	for (i = 0; i < 10; i++) {
		printf("-");
		sem_wait(&sem);
	}
	return NULL;
}

void *plus(void *argg) {
	unsigned int seed = *((unsigned int *) argg);
	int interval = RANDOM_WITHIN_RANGE(100000, 500000, seed);
	int i = 0;

	for (i = 0; i < 3; i++) {
		printf("+");
		usleep(interval);
		sem_post(&sem);
	}
	return NULL;
}

int main(int argc, char **argv) {
	setvbuf(stdout, NULL, _IONBF, 0);
	int i = 0;

	// Init Mutex
	pthread_mutex_init(&mutex, NULL);
	// Create thread pool and run threads
	pthread_t thread[2];
	pthread_create(&thread[i], NULL, &inc, NULL);
	pthread_create(&thread[i + 1], NULL, &dec, NULL);
	// Wait for threads to finish
	pthread_join(thread[i], NULL);
	pthread_join(thread[i + 1], NULL);
	// Destroy Mutex
	pthread_mutex_destroy(&mutex);
	// Print final value
	printf("%d\n", gv);

	// Init Semaphore
	sem_init(&sem, 0, 0);
	for (i = 0; i < 1; i++) {
		unsigned int *t = (unsigned int *) malloc(sizeof(unsigned int));
		*t = rand();

		pthread_create(&thread[i], NULL, &plus, (void *) t);
		pthread_create(&thread[i + 1], NULL, &minus, NULL);

		pthread_join(thread[i], NULL);
		pthread_join(thread[i + 1], NULL);
	}
	sem_destroy(&sem);

	return 0;
}