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


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