Thread: pthread newbie question

I just started reading a pthreads tutorial and I was wondering why this:

Code:

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

void *th_go (void *num) {
	printf("thread %d\n",*(int*)num);
	pthread_exit(NULL);
}

int main() {
	pthread_t th[3];
	int i;
	void *ptr=(void*)&i;
	for (i=0; i<3; i++) {
		pthread_create(&th[i],NULL,th_go,ptr);
	}	
	return 0;			
}

Does this:

Code:

[root~/C] ./a.out
thread 0
thread 1
thread 2
[root~/C] ./a.out
thread 1
thread 2
thread 2
[root~/C] ./a.out
thread 1
thread 1
[root~/C] ./a.out
thread 0
thread 1
thread 2
[root~/C] ./a.out
thread 1
thread 1
[root~/C] ./a.out
thread 0
thread 1
thread 2
[root~/C] ./a.out
thread 0
thread 2
thread 2

You are taking the address of i, and then changing it in another thread. How can you possibly expect it to have a consistent value from one call to another?

You could either cast the int into void *, or make sure that each pointer is unique (e.g. use malloc, or use an array with x elements, and pass the address of the relevant element to each thread).

--
Mats

Originally Posted by matsp

and then changing it in another thread.

What?!#?!? Where?

...oh because it's a race with the for() ? Still, what about this:

Code:

[root~/C] ./a.out
thread 1
thread 0
thread 2
thread 2

Hmmm...using pthread_exit() at the end of main seems to correct this.

[edit] All that does is prevent the occasional fourth thread (which I don't know why it would ever happen). It is a race in the for loop. This is not like fork()!

Yes, obviously, you pass the address of i to the thread, then continuing in main thread, you modify i. What value will you get when you read it?

Then you exit from the whole app without waiting for the threads to finish.

--
Mats

Originally Posted by matsp

Yes, obviously, you pass the address of i to the thread, then continuing in main thread, you modify i. What value will you get when you read it?

Then you exit from the whole app without waiting for the threads to finish.

Yep. But what leads to this (maybe one time in ten, without pthread_exit in main):

Code:

Creating 0
thread 0
Creating 1
Creating 2
thread 2
thread 0
thread 0       /* that's 4 */

I added the "Creating" line in main. I presume the last two end up with zero because it's a stack variable and main already ended?

So, what exactly are you asking?

--
Mats

Originally Posted by matsp

So, what exactly are you asking?

Maybe I am getting slightly side-tracked, but it's curiosity that often gets me thru this new stuff. The only case that's unclear to me is that last one -- am I right to believe i ends up as 0 after main ends? But mostly, where does #4 come from?

I can't explain where #4 comes from... The value of i after the loop should not be zero, but who knows if it's not being used....

--
Mats

Just like every malloc() should be paired with a free() - every call to pthread_create() should be paired with a call to either pthread_join() or pthread_detach(). The only exception is when you use a the PTHREAD_CREATE_DETACHED attribute.

>> using pthread_exit() at the end of main seems to ... prevent the occasional fourth thread
Well, in of itself, it prevents all the threads from being killed prematurely. Returning from main() is the same as calling exit() - which kills all running threads. pthread_exit() never returns, so you never return from main() and all other threads are allowed to run to completion.

>> am I right to believe i ends up as 0 after main ends?
Technically speaking, the behavior is undefined. To get an explanation for curiosity's sake, you would need to examine the generated assembly in context with the hardware it's running on (examine it in the debugger).

Even though accessing the same memory location by multiple threads without synchronization is "undefined" - a simple unsynchronized read of an address shouldn't cause problems - as long as that address is valid. In this case, after returning from main() or calling pthread_exit(), i is not longer valid.

gg

Originally Posted by Codeplug

Just like every malloc() should be paired with a free() - every call to pthread_create() should be paired with a call to either pthread_join() or pthread_detach(). The only exception is when you use a the PTHREAD_CREATE_DETACHED attribute.

That makes sense (I'm not positive I believe you completely yet). Thanks for the notice.

Originally Posted by MK27

That makes sense (I'm not positive I believe you completely yet). Thanks for the notice.

What part is it that you don't believe?

--
Mats

It's in the "RATIONALE" section for pthread_join and pthread_detach.
http://www.opengroup.org/onlinepubs/...read_join.html

gg

Originally Posted by matsp

Then you exit from the whole app without waiting for the threads to finish.

Technically I think that's okay, because a return from main() implies a pthread_join() on any threads which are still running. I'm not 100% sure though.

EDIT: Nope, I'm wrong.