helgrind/tests/tc15_laog_lockdel.c

#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>

/* Test that locks, having entered the lock acquisition tracking
   machinery, are forgotten by it when the client does
   pthread_{mutex,rwlock}_destroy.  2008-Nov-10: see comments below. */

int main ( void )
{
   int r;
   pthread_mutex_t *mx1, *mx2;
   assert (sizeof(pthread_mutex_t) <= 120);
   mx1 = malloc(120 + sizeof(pthread_mutex_t) - sizeof(pthread_mutex_t));
   mx2 = malloc(120 + sizeof(pthread_mutex_t) - sizeof(pthread_mutex_t));

   assert(mx1);
   assert(mx2);

   r = pthread_mutex_init( mx1, NULL ); assert(r==0);
   r = pthread_mutex_init( mx2, NULL ); assert(r==0);

   /* Establish order 1 -> 2 */
   fprintf(stderr, "Establish order 1 -> 2\n");
   r = pthread_mutex_lock( mx1 ); assert(r==0);
   r = pthread_mutex_lock( mx2 ); assert(r==0);

   r = pthread_mutex_unlock( mx1 ); assert(r==0);
   r = pthread_mutex_unlock( mx2 ); assert(r==0);

   /* Try order 2 -> 1.  This gives an error. */
   fprintf(stderr, "Try order 2 -> 1.  This gives an error.\n");
   r = pthread_mutex_lock( mx2 ); assert(r==0); /* error */
   r = pthread_mutex_lock( mx1 ); assert(r==0);

   r = pthread_mutex_unlock( mx1 ); assert(r==0);
   r = pthread_mutex_unlock( mx2 ); assert(r==0);

   /* De-initialise 2 and re-initialise it.  This gives it a new
      identity, so a second locking sequence 2 -> 1 should now be OK. */
   fprintf(stderr, 
           "Free 2 and re-allocate it.  This gives it a new identity,\n");
   fprintf(stderr, "so a second locking sequence 2 -> 1 should now be OK.\n");
   pthread_mutex_destroy( mx2 );



   r = pthread_mutex_init( mx2, NULL ); assert(r==0);

   r = pthread_mutex_lock( mx2 ); assert(r==0);
   r = pthread_mutex_lock( mx1 ); assert(r==0); /* no error */

   r = pthread_mutex_unlock( mx1 ); assert(r==0);
   r = pthread_mutex_unlock( mx2 ); assert(r==0);

   /* done */

   fprintf(stderr, "done\n");
   r = pthread_mutex_destroy( mx1 );
   r = pthread_mutex_destroy( mx2 );

   free( mx1 );
   free( mx2 );

   return 0;
}

/* 2008-Nov-10: I believe this test is flawed and requires further
   investigation.  I don't think it really tests what it claims to
   test.  In particular, it still gives the right results if
   "pthread_mutex_destroy( mx2 );" at line 46 is commented out.  In
   other words, laog somehow forgets about mx2 so that 2->1 lock
   sequence at lines 52/3 does not produce a complaint, EVEN WHEN the
   preceding "pthread_mutex_destroy( mx2 );" is not observed.  I don't
   know why this is, but it seems highly suspicious to me. */