Re: [PATCH v2 0/5] pid: add pidfd_open()

[Jann Horn]

On Sat, Mar 30, 2019 at 6:24 PM Linus Torvalds
<torvalds@xxxxxxxxxxxxxxxxxxxx> wrote:
> On Sat, Mar 30, 2019 at 10:12 AM Christian Brauner <christian@xxxxxxxxxx> wrote:
> > To clarify, what the Android guys really wanted to be part of the api is
> > a way to get race-free access to metadata associated with a given pidfd.
> > And the idea was that *if and only if procfs is mounted* you could do:
> >
> > int pidfd = pidfd_open(1234, 0);
> >
> > int procfd = open("/proc", O_RDONLY | O_CLOEXEC);
> > int procpidfd = ioctl(pidfd, PIDFD_TO_PROCFD, procfd);
>
> And my claim is that this is three system calls - one of them very
> hacky - to just do
>
>     int pidfd = open("/proc/%d", O_PATH);
>
> and you're done. It acts as the pidfd _and_ the way to get the
> associated status files etc.
>
> So there is absolutely zero advantage to going through pidfd_open().
>
> No. No. No.
>
> So the *only* reason for "pidfd_open()" is if you don't have /proc in
> the first place. In which case the whole PIDFD_TO_PROCFD is bogus.

So if, in the future, there is some sort of "create a new task and
return an fd to it" syscall, do you think it should always return
pidfds, or do you think it should return fds to /proc if procfs is
available? And if it should return fds to /proc, does that mean that
this "create a task" API should take an extra argument with a file
descriptor to the procfs instance you want to use?
(This can't always be implemented easily in userspace on top of normal
clone(), because if you create a task without a termination signal -
like a thread -, its PID can be recycled under you.)
An API like this would have less complexity stuffed into a single
syscall if it always returns pidfds, and if you then actually want an
fd to procfs, you can do the conversion that requires specifying a
procfs instance separately.

Of course, if you think that we shouldn't add an API for
pidfd-to-procfs conversion before we have an API for
clone()-with-an-fd-retval, that's understandable.