Question

Sorry just found this code here – http://www.yolinux.com/TUTORIALS/LinuxTutorialPosixThreads.html and the mutex was explained with this code, but it has gone a little over my head. I understand the function of the mutex, and that it protects a shared variable during its critical section. The specifics here are confusing me though! By my understanding, we are creating a new thread with pthread_create, it’s running the functionC process, which increments a counter. The counter is the protected variable, and, since both functions are running concurrently, the counter would return the wrong value if it was not protected by the mutex.

Is this correct/close ? Thanks a lot :).

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

void *functionC();
pthread_mutex_t mutex1 = PTHREAD_MUTEX_INITIALIZER;
int  counter = 0;

main()
{
   int rc1, rc2;
   pthread_t thread1, thread2;

   /* Create independent threads each of which will execute functionC */

   if( (rc1=pthread_create( &thread1, NULL, &functionC, NULL)) )
   {
      printf("Thread creation failed: %d\n", rc1);
   }

   if( (rc2=pthread_create( &thread2, NULL, &functionC, NULL)) )
   {
      printf("Thread creation failed: %d\n", rc2);
   }

   /* Wait till threads are complete before main continues. Unless we  */
   /* wait we run the risk of executing an exit which will terminate   */
   /* the process and all threads before the threads have completed.   */

   pthread_join( thread1, NULL);
   pthread_join( thread2, NULL); 

   exit(0);
}

void *functionC()
{
   pthread_mutex_lock( &mutex1 );
   counter++;
   printf("Counter value: %d\n",counter);
   pthread_mutex_unlock( &mutex1 );
}

Answer 1

If you don’t have the mutex in place, this is what might happen:

// initialization
counter = 0;

// thread 1 runs:
counter++;

// context switch
// thread 2 runs:
counter++;

// context switch
// thread 1 runs and printf "Counter value: 2"
printf("Counter value: %d\n",counter);

// context switch
// thread 2 runs and printf "Counter value: 2"
printf("Counter value: %d\n",counter);

So, you might end up with this output:

Counter value: 2
Counter value: 2

Now, with the mutex in place, you make sure that the increment and its print will run atomically, so you are 100% sure that the output will be:

Counter value: 1
Counter value: 2

But never:

Counter value: 2
Counter value: 2