Why are my threads I created not printed in order?

Question

I have this program:

void *func(void *arg) {
    pthread_mutex_lock(&mutex);
    int *id = (int *)arg;

    printf("My ID is %d\n" , *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?

Answer 1

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

int *id = (int *)arg;

printf("My ID is %d\n" , *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 %d\\n" , (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; } 

Answer 2

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 %d\n" , *id);