The existing docbook comments for the functions related to creating
a devicetree node do not explain the reference count of a newly
created node, how decrementing the reference count to zero will
free the associated memory, and the caller's responsibility to
call of_node_put() on the node. Explain what happens when the
reference count is decremented to zero.
Signed-off-by: Frank Rowand <frowand.list@gmail.com>
Link: https://lore.kernel.org/r/20230213185702.395776-8-frowand.list@gmail.com
Signed-off-by: Rob Herring <robh@kernel.org>
* another node. The node data are dynamically allocated and all the node
* flags have the OF_DYNAMIC & OF_DETACHED bits set.
*
- * Return: The newly allocated node or NULL on out of memory error.
+ * Return: The newly allocated node or NULL on out of memory error. Use
+ * of_node_put() on it when done to free the memory allocated for it.
*/
struct device_node *__of_node_dup(const struct device_node *np,
const char *full_name)
struct property *old_prop;
};
+/**
+ * of_node_init - initialize a devicetree node
+ * @node: Pointer to device node that has been created by kzalloc()
+ * @phandle_name: Name of property holding a phandle value
+ *
+ * On return the device_node refcount is set to one. Use of_node_put()
+ * on @node when done to free the memory allocated for it. If the node
+ * is NOT a dynamic node the memory will not be freed. The decision of
+ * whether to free the memory will be done by node->release(), which is
+ * of_node_release().
+ */
/* initialize a node */
extern const struct kobj_type of_node_ktype;
extern const struct fwnode_operations of_fwnode_ops;