sh-pfc: Don't take the sh_pfc spinlock in sh_pfc_map_gpios()
authorLaurent Pinchart <laurent.pinchart+renesas@ideasonboard.com>
Wed, 28 Nov 2012 19:56:48 +0000 (20:56 +0100)
committerLaurent Pinchart <laurent.pinchart+renesas@ideasonboard.com>
Fri, 15 Mar 2013 12:33:36 +0000 (13:33 +0100)
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>
drivers/pinctrl/sh-pfc/pinctrl.c

index 9bd0a830c140e78d21ea19335c7e7fa983f7437e..4ce2753cb2df03e52e2058c1743d7ec0537b28a7 100644 (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;
 }