Re: [PATCH] proc: fix return value of proc_reg_open() in "too late" case

[Alexey Dobriyan <adobriyan@gmail.com>]

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.