pinctrl: select_state: don't call pinctrl_free_setting on error
authorRichard Genoud <richard.genoud@gmail.com>
Fri, 29 Mar 2013 09:03:26 +0000 (10:03 +0100)
committerLinus Walleij <linus.walleij@linaro.org>
Wed, 3 Apr 2013 12:36:43 +0000 (14:36 +0200)
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: Richard Genoud <richard.genoud@gmail.com>
Reviewed-by: Stephen Warren <swarren@nvidia.com>
Signed-off-by: Linus Walleij <linus.walleij@linaro.org>
drivers/pinctrl/core.c

index 96cee79..deb3d04 100644 (file)
@@ -970,7 +970,15 @@ unapply_new_state:
        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) {