<div dir="ltr"><div><div>Completely agree with the flagpole analogy. In case of multiple references to a shared resource, the system does not consider the resource free as long as even one reference to it exists.<br><br></div>
About the race-condition though i am not so sure. The parent process is supposed to block on vfork() until the child process returns. This is a crucial difference between vfork() and the more common fork().<br><br><a href="http://linux.die.net/man/2/vfork" target="_blank">http://linux.die.net/man/2/vfork</a><br>
<a href="http://linux.die.net/man/2/fork" target="_blank">http://linux.die.net/man/2/fork</a><br><br>So in the above case i imagine the flow would be as follows:<br>1. Parent process starts execution.<br>2. vfork() called spawns child process.<br>
3. Child process executes(parent blocked at vfork).<br>4. Child process completes and exits.<br>5. Parent process continues.<br></div><div>6. Parent process completes and exits.<br><br></div><div>The additonal references to resources(memory for example) are created in step2 while spawning the child process. These references are deleted when step4 completes. But the original references created by the parent process in step1 still exist till end of step6 and hence resources will NOT be freed in the middle.<br>
</div><div><div><div><div><br></div><div>I believe the race condition is entirely plausible if fork() is used instead of vfork().<br><br></div><div>What interests me is the following line in the vfork manual page:<br> "...the behavior is undefined if the process created by
<span>vfork</span>() either modifies any data other than a variable of type <i>pid_t</i> used to store the return value from <span>vfork</span>()..."<br><br></div><div>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?...<br>
<br></div><div>regards<br></div><div>CVS<br></div><div><br>On Sat, Jan 19, 2013 at 3:19 AM, <<a href="mailto:Valdis.Kletnieks@vt.edu" target="_blank">Valdis.Kletnieks@vt.edu</a>> wrote:<br>><br>> On Fri, 18 Jan 2013 19:59:38 +0530, Niroj Pokhrel said:<br>
><br>> > I have been trying to create a process using vfork(). And both of the child<br>> > and the parent process execute it in the same address space. So, if I<br>> > execute exit(0) in the child process, it should throw some error right.<br>
><br>> Why do you think it should throw an error?<br>><br>> > Since the execution is happening in child process first and if I release<br>> > all the resources by using exit(0) in the child process then parent should<br>
> > be deprived of the resources and should throw some errors right ??<br>><br>> No, because those resources that were shared across a fork() or vfork() were in<br>> general *multiple references* to the same resource.<br>
><br>> As an example - imagine a flagpole. You grab it with your hand, you're<br>> now holding it. You invite your friend to come over and grab it with<br>> his hand - now he's holding it too.<br>><br>
> But either one of you can let go of the flagpole - and the other one is<br>> still holding the flagpole until *they* let go. And the order you let<br>> go doesn't matter in this case - which is important because your example<br>
> code has a race condition....<br>><br>> Note that there are other cases where the order people let go *does* matter.<br>> This is when you start having to worry about "locking order" and things like<br>
> that.<br>><br>> > In the following code, however the process ran fine even though I have<br>> > exit(0) in the child process ........<br>><br>> > #include<stdio.h><br>> > #include<stdlib.h><br>
> > #include<sys/types.h><br>> > #include<unistd.h><br>> > int main()<br>> > {<br>> > int val,i=0;<br>> > val=vfork();<br>> > if(val==0)<br>> > {<br>
> > printf("\nI am a child process.\n");<br>><br>> Note that printf() gets interesting due to stdio buffering. You probably<br>> want to call setbuf() and guarantee line-buffering of the output if you're<br>
> playing these sorts of games - the buffering can totally mask a real race<br>> condition or other bug.<br>><br>> > printf(" %d ",i++);<br>> > exit(0);<br>> > }<br>
> > else<br>> > {<br>><br>> /* race condition here - may want wait() or waitpid() to synchronize? */<br>><br>> > printf("\nI am a parent process.\n");<br>> > printf(" %d ",i);<br>
> > }<br>> > return 0;<br>> > }<br>> > // The program is running fine .....<br>> > But as I have read it should throw some error right ?? I don't know what I<br>> > am missing . Please point out the point I'm missing. Thanking you in<br>
> > advance.<br>><br>> You're also missing the fact that after the vfork(), there's no real<br>> guarantee of which will run first - which means that the parent can race<br>> and output the 'printf("%d",i)" *before* the child process gets a chance<br>
> to do the i++.<br>><br>> (Aside - for a while, there was a patch in place that ensured that the<br>> child would run first, on the theory that the child would often do something<br>> short that the parent was waiting on, so scheduling parent-first would just<br>
> result in the parent running, blocking to wait, and we end up running the<br>> child anyhow before the parent could continue. It broke an *amazing* amount<br>> of stuff in userspace because often the child would exit() before the parent was<br>
> ready to deal with the child process's termination. Usual failure mode was<br>> the parent would set a SIGCHLD handler, and wait for the signal which never<br>> happened because the SIGCHLD actually fired *before* the handler was set up).<br>
><br>> (And on non-cache-coherent systems, it's even possible that the i++ happens<br>> on a different CPU first, and the CPU running the parent process never becomes<br>> aware of it. See 'Documentation/memory-barriers.txt' in the Linux source<br>
> for more info on how this works for data inside the kernel. This example<br>> is out in userspace, so other techniques are required instead to do cross-CPU<br>> synchronization.<br>><br>> _______________________________________________<br>
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