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Filetage simple sans arguments

Cet exemple de base compte à des vitesses différentes sur deux threads que nous appelons sync (main) et async (new thread). Le thread principal compte à 15 à 1 Hz (1 s) tandis que le second compte à 10 à 0,5 Hz (2 s). Comme le thread principal se termine plus tôt, nous utilisons pthread_join pour le faire attendre à la fin de son async.

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

/* This is the function that will run in the new thread. */
void * async_counter(void * pv_unused) {
    int j = 0;
    while (j < 10) {
        printf("async_counter: %d\n", j);
        sleep(2);
        j++;
    }

    return NULL;
}

int main(void) {
    pthread_t async_counter_t;
    int i;
    /* Create the new thread with the default flags and without passing
     * any data to the function. */
    if (0 != (errno = pthread_create(&async_counter_t, NULL, async_counter, NULL))) {
        perror("pthread_create() failed");
        return EXIT_FAILURE;
    }

    i = 0;
    while (i < 15) {
        printf("sync_counter: %d\n", i);
        sleep(1);
        i++;
    }

    printf("Waiting for async counter to finish ...\n");

    if (0 != (errno = pthread_join(async_counter_t, NULL))) {
        perror("pthread_join() failed");
        return EXIT_FAILURE;
    }

    return EXIT_SUCCESS;
}

Copié ici: http://stackoverflow.com/documentation/c/3873/posix-threads/13405/simple-thread-without-arguments où il avait été initialement créé par M. Rubio-Roy .

Simple Mutex Usage

La bibliothèque de threads POSIX fournit l'implémentation de la primitive de mutex, utilisée pour l'exclusion mutuelle. Mutex est créé à l'aide de pthread_mutex_init et détruit à l'aide de pthread_mutex_destroy . L'obtention d'un mutex peut être effectuée à l'aide de pthread_mutex_lock ou pthread_mutex_trylock (selon le délai souhaité) et la libération d'un mutex s'effectue via pthread_mutex_unlock .

Un exemple simple utilisant un mutex pour sérialiser l’accès à la section critique suit. Tout d'abord, l'exemple sans utiliser un mutex. Notez que ce programme a une course de données en raison d'un accès non synchronisé à global_resource par les deux threads. Par conséquent, ce programme a un comportement indéfini :

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

// Global resource accessible to all threads
int global_resource;

// Threading routine which increments the resource 10 times and prints
// it after every increment
void* thread_inc (void* arg)
{
    for (int i = 0; i < 10; i++)
    {
        global_resource++;
        printf("Increment: %d\n", global_resource);
        // Make this thread slower, so the other one
        // can do more work
        sleep(1);
    }

    printf("Thread inc finished.\n");

    return NULL;
}

// Threading routine which decrements the resource 10 times and prints
// it after every decrement
void* thread_dec (void* arg)
{
    for (int i = 0; i < 10; i++)
    {
        global_resource--;
        printf("Decrement: %d\n", global_resource);
    }

    printf("Thread dec finished.\n");

    return NULL;
}

int main (int argc, char** argv)
{
    pthread_t threads[2];

    if (0 != (errno = pthread_create(&threads[0], NULL, thread_inc, NULL)))
    {
        perror("pthread_create() failed");
        return EXIT_FAILURE;
    }

    if (0 != (errno = pthread_create(&threads[1], NULL, thread_dec, NULL)))
    {
        perror("pthread_create() failed");
        return EXIT_FAILURE;
    }

    // Wait for threads to finish
    for (int i = 0; i < 2; i++)
    {
        if (0 != (errno = pthread_join(threads[i], NULL))) {
            perror("pthread_join() failed");
            return EXIT_FAILURE;
        }
    }

    return EXIT_SUCCESS;
}

Une sortie possible est:

Increment: 1
Decrement: 0
Decrement: -1
Decrement: -2
Decrement: -3
Decrement: -4
Decrement: -5
Decrement: -6
Decrement: -7
Decrement: -8
Decrement: -9
Thread dec finished.
Increment: -8
Increment: -7
Increment: -6
Increment: -5
Increment: -4
Increment: -3
Increment: -2
Increment: -1
Increment: 0
Thread inc finished.

Maintenant, si nous voulons synchroniser ces threads pour que nous voulions d'abord les incrémenter ou les décrémenter, puis les faire différemment, nous devons utiliser une primitive de synchronisation, telle que mutex:

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

// Global resource accessible to all threads
int global_resource;
// Mutex protecting the resource
pthread_mutex_t mutex;

// Threading routine which increments the resource 10 times and prints
// it after every increment
void* thread_inc (void* arg)
{
    // Pointer to mutex is passed as an argument
    pthread_mutex_t* mutex = arg;

    // Execute the following code without interrupts, all the way to the
    // point B
    if (0 != (errno = pthread_mutex_lock(mutex)))
    {
        perror("pthread_mutex_lock failed");
        exit(EXIT_FAILURE);
    }

    for (int i = 0; i < 10; i++)
    {
        global_resource++;
        printf("Increment: %d\n", global_resource);
        // Make this thread slower, so the other one
        // can do more work
        sleep(1);
    }

    printf("Thread inc finished.\n");

    // Point B:
    if (0 != (errno = pthread_mutex_unlock(mutex)))
    {
        perror("pthread_mutex_unlock failed");
        exit(EXIT_FAILURE);
    }

    return NULL;
}

// Threading routine which decrements the resource 10 times and prints
// it after every decrement
void* thread_dec (void* arg)
{
    // Pointer to mutex is passed as an argument
    pthread_mutex_t* mutex = arg;

    if (0 != (errno = pthread_mutex_lock(mutex)))
    {
        perror("pthread_mutex_lock failed");
        exit(EXIT_FAILURE);
    }

    for (int i = 0; i < 10; i++)
    {
        global_resource--;
        printf("Decrement: %d\n", global_resource);
    }

    printf("Thread dec finished.\n");

    // Point B:
    if (0 != (errno = pthread_mutex_unlock(mutex)))
    {
        perror("pthread_mutex_unlock failed");
        exit(EXIT_FAILURE);
    }

    return NULL;
}

int main (int argc, char** argv)
{
    pthread_t threads[2];
    pthread_mutex_t mutex;

    // Create a mutex with the default parameters
    if (0 != (errno = pthread_mutex_init(&mutex, NULL)))
    {
        perror("pthread_mutex_init() failed");
        return EXIT_FAILURE;
    }

    if (0 != (errno = pthread_create(&threads[0], NULL, thread_inc, &mutex)))
    {
        perror("pthread_create() failed");
        return EXIT_FAILURE;
    }

    if (0 != (errno = pthread_create(&threads[1], NULL, thread_dec, &mutex)))
    {
        perror("pthread_create() failed");
        return EXIT_FAILURE;
    }

    // Wait for threads to finish
    for (int i = 0; i < 2; i++)
    {
        if (0 != (errno = pthread_join(threads[i], NULL))) {
            perror("pthread_join() failed");
            return EXIT_FAILURE;
        }
    }

    // Both threads are guaranteed to be finished here, so we can safely
    // destroy the mutex
    if (0 != (errno = pthread_mutex_destroy(&mutex)))
    {
        perror("pthread_mutex_destroy() failed");
        return EXIT_FAILURE;
    }

    return EXIT_SUCCESS;
}

L'une des sorties possibles est

Increment: 1
Increment: 2
Increment: 3
Increment: 4
Increment: 5
Increment: 6
Increment: 7
Increment: 8
Increment: 9
Increment: 10
Thread inc finished.
Decrement: 9
Decrement: 8
Decrement: 7
Decrement: 6
Decrement: 5
Decrement: 4
Decrement: 3
Decrement: 2
Decrement: 1
Decrement: 0
Thread dec finished.

L'autre sortie possible serait inverse, au cas où thread_dec obtiendrait le mutex en premier.



Modified text is an extract of the original Stack Overflow Documentation
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