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