pinctrl: select_state: don't call pinctrl_free_setting on error

As Stephen Warren pointed out, pinctrl_free_setting() was called instead
of pinmux_disable_setting() on error.
In this error code, we want to call pinmux_disable_setting() where
pinmux_enable_setting() was called.
And when pinconf_apply_setting() was called, we can't do much to undo
the pin muxing (the closest thing I can think about for "unmuxing" a pin
is muxing it as GPIO input).
Signed-off-by: NRichard Genoud <richard.genoud@gmail.com>
Reviewed-by: NStephen Warren <swarren@nvidia.com>
Signed-off-by: NLinus Walleij <linus.walleij@linaro.org>

drivers/pinctrl/core.c

@@ -970,7 +970,15 @@ static int pinctrl_select_state_locked(struct pinctrl *p,
 	list_for_each_entry(setting2, &state->settings, node) {
 		if (&setting2->node == &setting->node)
 			break;
-		pinctrl_free_setting(true, setting2);
+		/*
+		 * All we can do here is pinmux_disable_setting.
+		 * That means that some pins are muxed differently now
+		 * than they were before applying the setting (We can't
+		 * "unmux a pin"!), but it's not a big deal since the pins
+		 * are free to be muxed by another apply_setting.
+		 */
+		if (setting2->type == PIN_MAP_TYPE_MUX_GROUP)
+			pinmux_disable_setting(setting2);
 	}
 
 	if (old_state) {