Re: [PATCH] BUG: [RFC] pinctrl: pins are freed 2 times in pinctrl_bind_pins
From: Stephen Warren
Date: Wed Mar 20 2013 - 12:24:00 EST
On 03/20/2013 05:31 AM, Richard Genoud wrote:
> If the function pinctrl_select_state() fails because one pin is already
> taken elsewhere, pinmux_enable_setting makes all the necessary pin_free
> calls (and not more than necessary).
> The problem here is that devm_pinctrl_put() will be called on the pin
> group, and each pin in this group has already been freed.
> If a i2c function has already sucessfully taken pins 5 and 6.
> And now, pinctrl_bind_pins() is called for function PHY (pins 3 4 5 6 7).
> pinmux_enable_setting() will fail AND call pin_free on necessary pins.
> But if devm_pinctrl_put() is called, it will call again pin_free on pins
> 3 4 5 6 7.
> So, the pins 5 and 6 will be released (and pins 3 4 7 double freed).
> Which means that even if the i2c function has claim the pins, they will
> be available for other functions.
> This patch simply doesn't call devm_pinctrl_put when
> pinctrl_select_state fails, but I'm not sure it's the right thing to do.
The correct fix here is not to skip the call to devm_pinctrl_put(),
since that undoes a lot of other things besides the current state selection.
Instead, pinctrl_select_state_locked() needs to be fixed so that:
Change "p->state = state;" to "p->state = NULL;" or similar, to indicate
that no state is selected. (Please validate if a NULL value in that
variable will cause problems elsewhere)
Add back the assignment "p->state = state;" at the end of the function,
if no error occurred.
Fix the list_for_each_entry() call that applies all the settings for the
new state so that if it fails, it undoes everything that it's applied so
far. That's the hard part, unless there's a
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