Re: [RFC] Per-thread getrusage

[Andrew Morton <akpm at linux-foundation dot org>]

On Mon, 28 Jan 2008 10:48:23 +0300 Pavel Emelyanov <xemul@openvz.org> wrote:

> Andrew Morton wrote:
> > 	On Thu, 17 Jan 2008 13:57:05 +0530 Vinay Sridhar <vinay@linux.vnet.ibm.com> wrote:
> > 
> >> Hi All,
> >>
> >> Last year, there was discussion about per-thread getrusage by adding
> >> RUSAGE_THREAD flag to getrusage(). Please refer to the thread
> >> http://lkml.org/lkml/2007/4/4/308. Ulrich had suggested that we should
> >> design a better user-space API. Specifically, we need a
> >> pthread_getrusage interface in the thread library, which accepts
> >> pthread_t, converts pthread_t into the corresponding tid and passes it
> >> down to the syscall.
> >>
> >> There are two ways to implement this in the kernel:
> >> 1) Introduce an additional parameter 'tid' to sys_getrusage() and put
> >> code in glibc to handle getrusage() and pthread_getrusage() calls
> >> correctly.
> >> 2) Introduce a new system call to handle pthread_getrusage() and leave
> >> sys_getrusage() untouched.
> >>
> >> We implemented the second idea above, simply because it avoids touching
> >> any existing code. We have implemented a new syscall, thread_getrusage()
> >> and we have exposed pthread_getrusage() API to applications.
> >>
> >> Could you please share your thoughts on this? Does the approach look
> >> alright? The code is hardly complete. It is just a prototype that works
> >> on IA32 at the moment.
> >>
> >> ...
> >>
> >> +asmlinkage long sys_thread_getrusage(int tid, struct rusage __user *ru);
> > 
> > What happens if `tid' refers to a thread in a different pid namespace?
> > 
> 
> That's impossible. I explicitly deny namespace creation in case the
> CLONE_THREAD is specified. So all threads of a single process always
> live in one pid namespace.
> 

If the code was using find_task_by_vpid() then OK (I guess).  But it is
looking the tids up in the init_pid_ns.  Which I assume means that if it's
in a new namespace and is looking up a sibling thread it will simply fail?

Or am I missing something?