I believe the race condition is entirely plausible if fork() is used instead of vfork().

What interests me is the following line in the vfork manual page:
"...the behavior is undefined if the process created by vfork() either modifies any data other than a variable of type pid_t used to store the return value from vfork()..."

So the behaviour is undefined for modifying the variable "i"? Does this mean that "i++" in the child process is NOT guaranteed to always reflect in the parent process? (Though in either case there still wont be any error/crash, just that the one outcome is not guaranteed across all platforms/archs/implementations). Is it so? Any real-world experiences of using vfork anyone?...

regards

CVS

On Sat, Jan 19, 2013 at 3:19 AM, <Valdis.Kletnieks@vt.edu> wrote:
>
> On Fri, 18 Jan 2013 19:59:38 +0530, Niroj Pokhrel said:
>
> > I have been trying to create a process using vfork(). And both of the child
> > and the parent process execute it in the same address space. So, if I
> > execute exit(0) in the child process, it should throw some error right.
>
> Why do you think it should throw an error?
>
> > Since the execution is happening in child process first and if I release
> > all the resources by using exit(0) in the child process then parent should
> > be deprived of the resources and should throw some errors right ??
>
> No, because those resources that were shared across a fork() or vfork() were in
> general *multiple references* to the same resource.
>
> As an example - imagine a flagpole.  You grab it with your hand, you're
> now holding it.  You invite your friend to come over and grab it with
> his hand - now he's holding it too.
>
> But either one of you can let go of the flagpole - and the other one is
> still holding the flagpole until *they* let go.  And the order you let
> go doesn't matter in this case - which is important because your example
> code has a race condition....
>
> Note that there are other cases where the order people let go *does* matter.
> This is when you start having to worry about "locking order" and things like
> that.
>
> > In the following code, however the process ran fine even though I have
> > exit(0) in the child process ........
>
> > #include<stdio.h>
> > #include<stdlib.h>
> > #include<sys/types.h>
> > #include<unistd.h>
> > int main()
> > {
> >     int val,i=0;
> >     val=vfork();
> >     if(val==0)
> >     {
> >         printf("\nI am a child process.\n");
>
> Note that printf() gets interesting due to stdio buffering.  You probably
> want to call setbuf() and guarantee line-buffering of the output if you're
> playing these sorts of games - the buffering can totally mask a real race
> condition or other bug.
>
> >         printf(" %d ",i++);
> >         exit(0);
> >     }
> >     else
> >     {
>
> /* race condition here - may want wait() or waitpid() to synchronize? */
>
> >         printf("\nI am a parent process.\n");
> >         printf(" %d ",i);
> >     }
> >     return 0;
> > }
> > // The program is running fine .....
> > But as I have read it should throw some error right ?? I don't know what I
> > am missing . Please point out the point I'm missing. Thanking you in
> > advance.
>
> You're also missing the fact that after the vfork(), there's no real
> guarantee of which will run first - which means that the parent can race
> and output the 'printf("%d",i)" *before* the child process gets a chance
> to do the i++.
>
> (Aside - for a while, there was a patch in place that ensured that the
> child would run first, on the theory that the child would often do something
> short that the parent was waiting on, so scheduling parent-first would just
> result in the parent running, blocking to wait, and we end up running the
> child anyhow before the parent could continue.  It broke an *amazing* amount
> of stuff in userspace because often the child would exit() before the parent was
> ready to deal with the child process's termination. Usual failure mode was
> the parent would set a SIGCHLD handler, and wait for the signal which never
> happened because the SIGCHLD actually fired *before* the handler was set up).
>
> (And on non-cache-coherent systems, it's even possible that the i++ happens
> on a different CPU first, and the CPU running the parent process never becomes
> aware of it.  See 'Documentation/memory-barriers.txt' in the Linux source
> for more info on how this works for data inside the kernel.  This example
> is out in userspace, so other techniques are required instead to do cross-CPU
> synchronization.
>
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