Multithreaded program with mutex on mutual resource [duplicate] - c

Question

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Pthread_create() incorrect start routine parameter passing
(1 answer)
Closed 3 years ago.
I tried to build a program which should create threads and assign a Print function to each one of them, while the main process should use printf function directly.
Firstly, I made it without any synchronization means and expected to get a randomized output.
Later I tried to add a mutex to the Print function which was assigned to the threads and expected to get a chronological output but it seems like the mutex had no effect about the output.
Should I use a mutex on the printf function in the main process as well?
Thanks in advance
My code:
#include <stdio.h>
#include <pthread.h>
#include <errno.h>
pthread_t threadID[20];
pthread_mutex_t lock;
void* Print(void* _num);
int main(void)
{
int num = 20, indx = 0, k = 0;
if (pthread_mutex_init(&lock, NULL))
{
perror("err pthread_mutex_init\n");
return errno;
}
for (; indx < num; ++indx)
{
if (pthread_create(&threadID[indx], NULL, Print, &indx))
{
perror("err pthread_create\n");
return errno;
}
}
for (; k < num; ++k)
{
printf("%d from main\n", k);
}
indx = 0;
for (; indx < num; ++indx)
{
if (pthread_join(threadID[indx], NULL))
{
perror("err pthread_join\n");
return errno;
}
}
pthread_mutex_destroy(&lock);
return 0;
}
void* Print(void* _indx)
{
pthread_mutex_lock(&lock);
printf("%d from thread\n", *(int*)_indx);
pthread_mutex_unlock(&lock);
return NULL;
}

Answer 1

All questions of program bugs notwithstanding, pthreads mutexes provide only mutual exclusion, not any guarantee of scheduling order. This is typical of mutex implementations. Similarly, pthread_create() only creates and starts threads; it does not make any guarantee about scheduling order, such as would justify an assumption that the threads reach the pthread_mutex_lock() call in the same order that they were created.
Overall, if you want to order thread activities based on some characteristic of the threads, then you have to manage that yourself. You need to maintain a sense of which thread's turn it is, and provide a mechanism sufficient to make a thread notice when it's turn arrives. In some circumstances, with some care, you can do this by using semaphores instead of mutexes. The more general solution, however, is to use a condition variable together with your mutex, and some shared variable that serves as to indicate who's turn it currently is.

Answer 2

The code passes the address of the same local variable to all threads. Meanwhile, this variable gets updated by the main thread.
Instead pass it by value cast to void*.
Fix:
pthread_create(&threadID[indx], NULL, Print, (void*)indx)
// ...
printf("%d from thread\n", (int)_indx);
Now, since there is no data shared between the threads, you can remove that mutex.

Answer 3

All the threads created in the for loop have different value of indx. Because of the operating system scheduler, you can never be sure which thread will run. Therefore, the values printed are in random order depending on the randomness of the scheduler. The second for-loop running in the parent thread will run immediately after creating the child threads. Again, the scheduler decides the order of what thread should run next.
Every OS should have an interrupt (at least the major operating systems have). When running the for-loop in the parent thread, an interrupt might happen and leaves the scheduler to make a decision of which thread to run. Therefore, the numbers being printed in the parent for-loop are printed randomly, because all threads run "concurrently".
Joining a thread means waiting for a thread. If you want to make sure you print all numbers in the parent for loop in chronological order, without letting child thread interrupt it, then relocate the for-loop section to be after the thread joining.