On Mon, Oct 26, 2009 at 11:57:29AM -0500, Serge E. Hallyn wrote:
> Quoting Jan Kara (jack@xxxxxxx):
> > Hi,
> >
> > On Sun 25-10-09 07:29:53, Pavel Machek wrote:
> > > ...yes, they do exist, in /proc/self/fd/* . Unfortunately, their
> > > permissions are not actually checked during open, resulting in
> > > (obscure) security hole: if you have fd open for reading, you can
> > > reopen it for write, even through unix permissions would not allow
> > > that.
> > >
> > > Now... I'd like to close the hole. One way would be to actually check
> > > symlink permissions on open -- because those symlinks already have
> > > correct permissions.
> > Hmm, I'm not sure I understand the problem. Symlink is just a file
> > containing a path. So if you try to open a symlink, you will actually open
> > a file to which the path points. So what security problem is here? Either
> > you can open the file symlink points to for writing or you cannot...
> > Anyway, if you want to play with this,
> > fs/proc/base.c:proc_pid_follow_link
> > is probably the function you are interested in.
>
> The problem he's trying to address is that users may try to protect
> a file by doing chmod 700 on the parent dir, but leave the file itself
> accessible. They don't realize that merely having a task with an open
> fd to that file gives other users another path to the file.
>
> Whether or not that's actually a problem is open to debate, but I think
> he's right that many users aren't aware of it.
If /proc/self/fd/23 is a symlink to /home/me/privatedir/secret, then an
open("proc/self/fd/23",...) still traverses the whole /home/.../secret
path, and needs appropriate permissions at each step, doesn't it?
Probably I'm just terminally confused....
--b.
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