Re: [PATCH] of: Add missing of_node_put() in of_find_node_by_path()

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Hi Grant,

On Thu, Jan 22, 2015 at 5:14 PM, Grant Likely <grant.likely@xxxxxxxxxx> wrote:
> On Wed, 14 Jan 2015 16:45:56 +0100
> , Geert Uytterhoeven <geert+renesas@xxxxxxxxx>
>  wrote:
>> When traversing all nodes and moving to a new path component, the old
>> one must be released by calling of_node_put(). Else the refcounts of the
>> parent node(s) will not be decremented.
>>
>> Signed-off-by: Geert Uytterhoeven <geert+renesas@xxxxxxxxx>
>> ---
>> Background.
>>
>> While investigating a reference count imbalance issue with
>> CONFIG_OF_DYNAMIC=y, I wrote the debug code below to validate the
>> reference counts of the nodes I was interested in.
>> During the first call of check_refcnts(), it gathers all reference
>> counts. During a subsequent call, it verifies that they are still the
>> same.
>>
>> Surprisingly, lots of reference counts were wrong, and kept incrementing
>> every time check_refcnts() was called.

>> diff --git a/drivers/of/base.c b/drivers/of/base.c
>> index 36536b6a8834acd2..f3e346e19c69d1f2 100644
>> --- a/drivers/of/base.c
>> +++ b/drivers/of/base.c
>> @@ -791,8 +791,10 @@ struct device_node *of_find_node_opts_by_path(const char *path, const char **opt
>>       if (!np)
>>               np = of_node_get(of_root);
>>       while (np && *path == '/') {
>> +             struct device_node *parent = np;
>>               path++; /* Increment past '/' delimiter */
>> -             np = __of_find_node_by_path(np, path);
>> +             np = __of_find_node_by_path(parent, path);
>> +             of_node_put(parent);
>>               path = strchrnul(path, '/');
>>       }
>>       raw_spin_unlock_irqrestore(&devtree_lock, flags);
>
> This doesn't look /quite/ the best. __for_each_child_of_node() is
> fiddling with refcounts, but the '__' of functions shouldn't need to do
> that since they are called under the spinlock (nothing is going to
> change while they are called). __of_find_all_nodes() for instance
> doesn't do refcounting, but of_find_all_nodes() does.
>
> Does the following also solve the problem?

Yes, it does.

Tested-by: Geert Uytterhoeven <geert+renesas@xxxxxxxxx>

> diff --git a/drivers/of/base.c b/drivers/of/base.c
> index 36536b6a8834..0357b51a7440 100644
> --- a/drivers/of/base.c
> +++ b/drivers/of/base.c
> @@ -626,9 +626,8 @@ static struct device_node *__of_get_next_child(const struct device_node *node,
>
>         next = prev ? prev->sibling : node->child;
>         for (; next; next = next->sibling)
> -               if (of_node_get(next))
> +               if (next)
>                         break;
> -       of_node_put(prev);
>         return next;
>  }
>  #define __for_each_child_of_node(parent, child) \
> @@ -650,7 +649,8 @@ struct device_node *of_get_next_child(const struct device_node *node,
>         unsigned long flags;
>
>         raw_spin_lock_irqsave(&devtree_lock, flags);
> -       next = __of_get_next_child(node, prev);
> +       next = of_node_get(__of_get_next_child(node, prev));
> +       of_node_put(prev);
>         raw_spin_unlock_irqrestore(&devtree_lock, flags);
>         return next;
>  }
> @@ -789,12 +789,13 @@ struct device_node *of_find_node_opts_by_path(const char *path, const char **opt
>         /* Step down the tree matching path components */
>         raw_spin_lock_irqsave(&devtree_lock, flags);
>         if (!np)
> -               np = of_node_get(of_root);
> +               np = of_root;
>         while (np && *path == '/') {
>                 path++; /* Increment past '/' delimiter */
>                 np = __of_find_node_by_path(np, path);
>                 path = strchrnul(path, '/');
>         }
> +       of_node_get(np);
>         raw_spin_unlock_irqrestore(&devtree_lock, flags);
>         return np;
>  }

Gr{oetje,eeting}s,

                        Geert

--
Geert Uytterhoeven -- There's lots of Linux beyond ia32 -- geert@xxxxxxxxxxxxxx

In personal conversations with technical people, I call myself a hacker. But
when I'm talking to journalists I just say "programmer" or something like that.
                                -- Linus Torvalds
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