Ytukyg

I have this program:

void *func(void *arg) {

    pthread_mutex_lock(&mutex);

    int *id = (int *)arg;



    printf("My ID is %dn" , *id);

    pthread_mutex_unlock(&mutex);

}



int main() {

    int i;

    pthread_t tid[3];



    // Let us create three threads

    for (i = 0; i < 3; i++) {

        pthread_create(&tid[i], NULL, func, (void *)&i);

    }



    for (i = 0; i < 3; i++) {

        pthread_join(tid[i], NULL);

    }



    pthread_exit(NULL);

    return 0;

}

I expected it to output this:

My ID is 0

My ID is 1

My ID is 2

But instead I get random output, such as this:

My ID is 0

My ID is 0

My ID is 2

Since I already added mutex lock, I thought it would solve the problem. What else did I do wrong? Is this related to race condition?

Here id points to the same variable i in main for all the threads.

int *id = (int *)arg;



printf("My ID is %dn" , *id);

But the variable i is constantly being update by the two for-loops in main behind the threads back. So before the thread reaches the point of printf, the value of i, and therefore also the value of *id, may have changed.

There are a few ways to solve this. The best way depends on the use case:

1. Wait in main until the thread signals that it has made a copy of *id before modifying i or letting it go out of scope.


2. Declare and initialize an array, int thread_id, and create the threads like this:
   pthread_create(&tid[i], NULL, func, &thread_id[i]);
   


3. 
   

   malloc some memory and and initialize it with a copy of i:

   
   
   

   int *thread_id = malloc(sizeof(*thread_id));
   
   *thread_id = i
   
   pthread_create(&tid[i], NULL, func, thread_id);
   
   

   
   
   

   Just don't forget to free your memory int the thread when you are finished using it. Or in main if the thread fails to start.

   
   


4. 
   

   If i fits in a void * can pass its content directly as a parameter to the thread. To make sure it fits, you can declare it as intptr_t rather than int
   (We basicly abuse the fact that pointers are nothing more than magic integers) :

   
   
   

   void *func(void *arg) {
   
       pthread_mutex_lock(&mutex);
   
       // Here we interpret a pointer value as an integer value
   
       intptr_t id = (intptr_t )arg;
   
   
   
       printf("My ID is %dn" , (int)id);
   
       pthread_mutex_unlock(&mutex);
   
   }
   
   
   
   int main() {
   
       intptr_t i;
   
       pthread_t tid[3];
   
   
   
       // Let us create three threads
   
       for (i = 0; i < 3; i++) {
   
           // Here we squeeze the integer value of `i` into something that is
   
           // supposed to hold a pointer
   
           pthread_create(&tid[i], NULL, func, (void *)i);
   
       }
   
   
   
       for (i = 0; i < 3; i++) {
   
           pthread_join(tid[i], NULL);
   
       }
   
   
   
       // This does not belong here !!
   
       // pthread_exit(NULL);
   
       return 0;
   
   }
   
   

   
   

Nope, no race conditions involved. (my b) There can be a race condition on i because all threads access it. Each thread gets started with a pointer to i. However, the main problem is that there is no guarantee that the thread will start and run the critical section while i holds the value you expect, in an order that you expect.

I'm assuming you declared the variable mutex globally and called pthread_mutex_init() somewhere to initialize it.

Mutexes are great to allow only one thread to access a critical section of code at a time. So the code as you've written creates all three threads to run in parallel, but only lets one thread at a time run the following code.

int *id = (int *)arg;



printf("My ID is %dn" , *id);