Re: [PATCH 2/6] Driver core: Fix memory leak on bus_register error path

[Patrick Pannuto <ppannuto@codeaurora.org>]

>>>  
>>>  	retval = kset_register(&priv->subsys);
>>> -	if (retval)
>>> +	if (retval) {
>>> +		kfree(priv->subsys.kobj.name);
>>
>> I don't think we want to bury the logic of how kobject names are handled
>> up here in the bus code, right?  Shouldn't the subsys kobject name be
>> able to be cleaned up on its own somehow instead?
>>
> 
> So, my first instinct was to use kobject_cleanup, but a few lines above:
> 
> 	priv->subsys.kobj.ktype = &bus_ktype;
> 
> and bus_ktype's definition, with the notable absence of a release method:
> 
> static struct kobj_type bus_ktype = {
> 	.sysfs_ops	= &bus_sysfs_ops,
> };
> 
> which in kobject_cleanup would yield:
> 
> 	struct kobj_type *t = get_ktype(kobj);
> 
> 	if (t && !t->release)
> 		pr_debug("kobject: '%s' (%p): does not have a release() "
> 			 "function, it is broken and must be fixed.\n",
> 			 kobject_name(kobj), kobj);
> 
> (if I understand everything correctly)
> 

Thinking about this more, how does this ever work "correctly"? If we
chase the 'normal' path from bus_unregister...

given bus->p->subsys.kobj.ktype = &bus_ktype

bus_unregister(struct bus_type *bus) {
   kset_unregister(&bus->p->subsys) {
      kobject_put(&k->kobj) {
         kref_put(&kobj->kref, kobject_release) {
            (assuming last ref)
            kobject_cleanup(container_of(kref, struct kobject, kref)) {
               	struct kobj_type *t = get_ktype(kobj);

		if (t && !t->release)
			pr_debug("kobject: '%s' (%p): does not have a release() "
				 "function, it is broken and must be fixed.\n",
				 kobject_name(kobj), kobj);

Wouldn't this get hit every time a bus unregisters?  I feel like I'm
missing something here?