Why are my threads I created not printed in order? - c

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