Re: pipe_release oops.

[Al Viro]

On Mon, Mar 11, 2013 at 08:10:10AM -0700, Linus Torvalds wrote:
> On Sun, Mar 10, 2013 at 5:35 PM, Al Viro <viro@xxxxxxxxxxxxxxxxxx> wrote:
> >
> > Hmm...  How the devil would things like pipe_read_open() get called, anyway?
> > pipe_rdwr_open() can be called, all right - that happens if you do pipe()
> > and then open() via /proc/self/fd/<n>.  But how could pipe_read_open() and
> > pipe_write_open() be called?  They are accessible only as ->open() of
> > read_pipefifo_fops/write_pipefifo_fops.  And those are only used by
> > fifo_open(), which does reassign file->f_op to them, but does *not* call
> > their ->open()...
> 
> .. same deal? Open the resulting fd from /proc/self/fd/<n> afterwards,
> which will call file->f_op->open(), no?

Not really - that would call inode->i_fop, not file->f_op.  You get dentry
of opened file, but file->f_op is set from scratch - not copied from the
original struct file.
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