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SubjectRe: [PATCH] proc: fix return value of proc_reg_open() in "too late" case
On Tue, Sep 02, 2008 at 04:26:24PM -0700, Andrew Morton wrote:
> On Sat, 30 Aug 2008 09:34:12 +0400
> Alexey Dobriyan <adobriyan@gmail.com> wrote:
>
> > If ->open() wasn't called, returning 0 is misleading and, theoretically,
> > oopsable:
> > 1. remove_proc_entry clears ->proc_fops, drops lock,
> > 2. ->open "succeeds",
> > 3. ->release oopses, because it assumes ->open was called (single_release()).
> >
> > Signed-off-by: Alexey Dobriyan <adobriyan@gmail.com>
> > ---
> >
> > fs/proc/inode.c | 2 +-
> > 1 file changed, 1 insertion(+), 1 deletion(-)
> >
> > --- a/fs/proc/inode.c
> > +++ b/fs/proc/inode.c
> > @@ -350,7 +350,7 @@ static int proc_reg_open(struct inode *inode, struct file *file)
> > if (!pde->proc_fops) {
> > spin_unlock(&pde->pde_unload_lock);
> > kfree(pdeo);
> > - return rv;
> > + return -EINVAL;
> > }
> > pde->pde_users++;
> > open = pde->proc_fops->open;
>
> Can this code path ever actually be executed? afacit if ->proc_fops is
> ever NULL, the caller (proc_get_inode) would have already oopsed:
>
> #ifdef CONFIG_COMPAT
> if (!de->proc_fops->compat_ioctl)
> inode->i_fop =
> &proc_reg_file_ops_no_compat;
> else
> #endif
> inode->i_fop = &proc_reg_file_ops;

Yes, it can.

remove_proc_entry() clears ->proc_fops to indicate that removal of PDE started.
But somebody could have already open(2) and is currently at the beginning of
proc_reg_open().

Well, that's how bug was noticed in the first place.



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