sh-pfc: Don't take the sh_pfc spinlock in sh_pfc_map_gpios()

The sh_pfc_map_gpios() function is only called at initialization time
when no other task can access the sh_pfc fields. Don't protect the
operation with a spinlock.

Signed-off-by: Laurent Pinchart <laurent.pinchart+renesas@ideasonboard.com>
Acked-by: Linus Walleij <linus.walleij@linaro.org>
This commit is contained in:
Laurent Pinchart 2012-11-28 20:56:48 +01:00
parent e3e89ae43e
commit d785fdb5d8

View File

@ -334,7 +334,6 @@ static void sh_pfc_map_one_gpio(struct sh_pfc *pfc, struct sh_pfc_pinctrl *pmx,
/* pinmux ranges -> pinctrl pin descs */
static int sh_pfc_map_gpios(struct sh_pfc *pfc, struct sh_pfc_pinctrl *pmx)
{
unsigned long flags;
int i;
pmx->nr_pads = pfc->info->last_gpio - pfc->info->first_gpio + 1;
@ -346,8 +345,6 @@ static int sh_pfc_map_gpios(struct sh_pfc *pfc, struct sh_pfc_pinctrl *pmx)
return -ENOMEM;
}
spin_lock_irqsave(&pfc->lock, flags);
/*
* We don't necessarily have a 1:1 mapping between pin and linux
* GPIO number, as the latter maps to the associated enum_id.
@ -368,8 +365,6 @@ static int sh_pfc_map_gpios(struct sh_pfc *pfc, struct sh_pfc_pinctrl *pmx)
sh_pfc_map_one_gpio(pfc, pmx, gpio, i);
}
spin_unlock_irqrestore(&pfc->lock, flags);
return 0;
}