static pthread_mutex_t my_mutex
- pourquoi doit on declarer pthread_mutex_t en static dans le code suivant (dans les premières lignes)
- que se passe t -il si on enlève static, quels sont les risques encourus ?
merci d'avance
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#include <stdlib.h>
#include <pthread.h>
static pthread_mutex_t my_mutex;
static int tab[5];
void *read_tab_process (void * arg)
{
int i;
pthread_mutex_lock (&my_mutex);
for (i = 0 ; i != 5 ; i++)
printf ("read_process, tab[%d] vaut %d\n", i, tab[i]);
pthread_mutex_unlock (&my_mutex);
pthread_exit (0);
}
void *write_tab_process (void * arg)
{
int i;
pthread_mutex_lock (&my_mutex);
for (i = 0 ; i != 5 ; i++) {
tab[i] = 2 * i;
printf ("write_process, tab[%d] vaut %d\n", i, tab[i]);
sleep (1); /* Relentit le thread d'ecriture... */
}
pthread_mutex_unlock (&my_mutex);
pthread_exit (0);
}
main (int ac, char **av)
{
pthread_t th1, th2;
void *ret;
pthread_mutex_init (&my_mutex, NULL);
if (pthread_create (&th1, NULL, write_tab_process, NULL) < 0) {
fprintf (stderr, "pthread_create error for thread 1\n");
exit (1);
}
if (pthread_create (&th2, NULL, read_tab_process, NULL) < 0) {
fprintf (stderr, "pthread_create error for thread 2\n");
exit (1);
}
(void)pthread_join (th1, &ret);
(void)pthread_join (th2, &ret);
} |
