Messages in this thread | | | From | Geert Uytterhoeven <> | Date | Thu, 2 Mar 2023 08:49:44 +0100 | Subject | Re: [PATCH 2/3] of: irq: make callers of of_irq_parse_one() release the device node |
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Hi Jean-Jacques,
Thanks for your patch!
On Wed, Mar 1, 2023 at 7:53 PM Jean-Jacques Hiblot <jjhiblot@traphandler.com> wrote: > of_irq_parse_one() does a get() on the device node returned in out_irq->np. > Callers of of_irq_parse_one() must do a put() when they are done with it.
What does "be done with it" really mean here?
> Signed-off-by: Jean-Jacques Hiblot <jjhiblot@traphandler.com>
> --- a/arch/arm/mach-shmobile/regulator-quirk-rcar-gen2.c > +++ b/arch/arm/mach-shmobile/regulator-quirk-rcar-gen2.c > @@ -184,6 +184,7 @@ static int __init rcar_gen2_regulator_quirk(void) > kfree(quirk); > continue; > } > + of_node_put(argsa->np);
The quirk object, which is a container of argsa, is still used below, and stored in a linked list. I agree argsa->np is not dereferenced, but the pointer itself is still compared to other pointers. IIUIC, calling of_node_put() might cause the reference count to drop to zero, and the underlying struct node object to be deallocated. So when a future reference to the same DT node will be taken, a new struct node object will be allocated, and the pointer comparison below will fail?
Or am I missing something?
Gr{oetje,eeting}s,
Geert
-- Geert Uytterhoeven -- There's lots of Linux beyond ia32 -- geert@linux-m68k.org
In personal conversations with technical people, I call myself a hacker. But when I'm talking to journalists I just say "programmer" or something like that. -- Linus Torvalds
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