On Mon, Mar 25, 2019 at 9:40 PM Jonathan Kowalski [off-list ref] wrote:

On Mon, Mar 25, 2019 at 8:34 PM Jann Horn [off-list ref] wrote:

quoted

 [...SNIP...]

Please don't do that. /proc/$pid/fd refers to the set of file
descriptors the process has open, and semantically doesn't have much
to do with the identity of the process. If you want to have a procfs
directory entry for getting a pidfd, please add a new entry. (Although
I don't see the point in adding a new procfs entry for this when you
could instead have an ioctl or syscall operating on the procfs
directory fd.)

There is no new entry. What I was saying (and I should have been
clearer) is that the existing entry for the fd when open'd with
O_DIRECTORY makes the kernel resolve the symlink to /proc/<PID> of the
process it maps to, so it would become:

  int dirfd = open("/proc/self/fd/3", O_DIRECTORY|O_CLOEXEC);

That still seems really weird. This magically overloads O_DIRECTORY,
which means "fail if the thing is not a directory", to suddenly have
an entirely different meaning for one magical special type of file. On
top of that, unlike an ioctl or a new syscall, it doesn't convey
explicit intent and increases the risk of confused deputy issues.

This also means you cannot cross the filesystem boundry, the said
process needs to have a visible entry (which would mean hidepid= and
gid= based access controls are honored), and you can only open the
dirfd of a process in the current ns (as the PID will not map to an
existent process if the pidfd maps to a process not in the same or
children pid ns, in fdinfo it lists -1 in the pid field (we might not
even need fdinfo anymore)).

AFAICS that doesn't have anything to do with whether you do this as a
syscall, as an ioctl, or as a jumped symlink. The kernel would have to
do the same security checks in any of those cases - only a classic,
non-jumped symlink would implicitly go through the existing permission
checks. And if you implement this with a non-jumped symlink, you get
races.