pthreads newby issues

[santiagorf]

Hi,
I have run this example, and I'm not getting the expected result.

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

    #define it 2

    pthread_t threads[it];

    pthread_mutex_t mu;

    void *thread_func(void *arg){
      int i = *(int *) arg;
      pthread_mutex_lock(&mu);
      printf("thread---  %d\n", i);

      pthread_mutex_unlock(&mu);
    }

    void main(){
      int i;
      pthread_mutex_init(&mu,NULL);


      int vec[it];
      for(i=0; i<it; i++)
        vec[i]=i;

      for (i=0; i<it; i++)
        pthread_create(&(threads[i]), NULL, thread_func, &i);


      for (i=0; i<it; i++)
        pthread_join(threads[i], NULL);

    }

When I run this code, it shows me random results. For example, I can get

thread--- 1
thread--- 0

or
thread--- 0
thread--- 0

I tried locking the printf function, as shown in the example, to no avail. But if I change the line

       pthread_create(&(threads[i]), NULL, thread_func, &i);

by this

        pthread_create(&(threads[i]), NULL, thread_func, &vec[i]);

I always get the intended result (even with locks removes) that is

thread--- 1
thread--- 0

or this

thread--- 0
thread--- 1

What is the problem with the original code?

thanks in advance!

[Peter87]

When 11 is executed it will read the value of i from main, but you don't know when this will happen so you don't know what the value is going to be.

[santiagorf]

This is what I thought, but since i is assigned before the thread is created i still found the output

thread--- 0
thread--- 0

strange.

[Peter87]

It's not strange. You set i to 0 again in the last loop.

[santiagorf]

I see. Thanks!!

[jared181]

Hi,
That is a nice thread.

I have a question, where is the mutex being locked by each thread?

I only see it getting unlocked by each thread.

Thank you.

BTW, where did you gain that Linux knowledge?

I'm trying to get into Operating Systems world, and into Linux specifically.