On Tue, Jul 22, 2014 at 05:25:03PM +0000, Jeff Haran wrote: > > -----Original Message----- > > From: Greg KH [mailto:greg@xxxxxxxxx] > > Sent: Monday, July 21, 2014 7:18 PM > > To: Jeff Haran > > Cc: kernelnewbies@xxxxxxxxxxxxxxxxx > > Subject: Re: question about kref API > > > > On Tue, Jul 22, 2014 at 12:27:20AM +0000, Jeff Haran wrote: > > > Hi, > > > > > > I've been reading Documentation/kref.txt in order to understand the > > > usage of this API. There is part of this documentation that I am > > > having difficulty understanding and was hoping somebody on this list > > > could clarify this. One of my assumptions in reading this is that the > > > expected usage of this API is that for a given object embedding a > > > kref, once the object has been initialized the number of calls to > > > "put" a given instance of an object should never exceed the number of > > > calls to "get" that same instance. > > > > If it does, the object will be cleaned up and deleted from the system, > > so you no longer have a valid pointer. > > > > > Maybe that's the root of my misunderstanding, but my assumption is > > > that calls to kref_get() and kref_put() are expected to be made in > > > pairs for a given instance of struct kref. If this is wrong, please > > > let me know. > > > > "pairs" in that the same number must be called in order for things to > > work properly. > > > > > Kref.txt includes some sample code that discusses using a mutex to > > > serialize the execution of its get_entry() and put_entry() functions: > > > > > > 146 static DEFINE_MUTEX(mutex); > > > 147 static LIST_HEAD(q); > > > 148 struct my_data > > > 149 { > > > 150 struct kref refcount; > > > 151 struct list_head link; > > > 152 }; > > > 153 > > > 154 static struct my_data *get_entry() > > > 155 { > > > 156 struct my_data *entry = NULL; > > > 157 mutex_lock(&mutex); > > > 158 if (!list_empty(&q)) { > > > 159 entry = container_of(q.next, struct my_data, link); > > > 160 kref_get(&entry->refcount); > > > 161 } > > > 162 mutex_unlock(&mutex); > > > 163 return entry; > > > 164 } > > > 165 > > > 166 static void release_entry(struct kref *ref) > > > 167 { > > > 168 struct my_data *entry = container_of(ref, struct my_data, refcount); > > > 169 > > > 170 list_del(&entry->link); > > > 171 kfree(entry); > > > 172 } > > > 173 > > > 174 static void put_entry(struct my_data *entry) > > > 175 { > > > 176 mutex_lock(&mutex); > > > 177 kref_put(&entry->refcount, release_entry); > > > 178 mutex_unlock(&mutex); > > > 179 } > > > > > > The sample code does not show the creation of the link list headed by > > > q, > > > > That is done there in the static initializer. [meta comment, please properly wrap your lines...] > You are referring to line 147 here, right? That creates an empty list > if I am following the code correctly. Yes. > What I meant was that the sample code doesn't show how instances of > struct my_data get initialized and inserted into the list at q. Makes > sense to leave that out for brevity I suppose and you've made it clear > below that for this to work right a call to kref_init() must have been > made on the refcount fields of any struct my_data instances that got > put into the list headed by q. Thanks. Yes, that code is left as an exercise for the reader. Or you can just look at the kernel where it is used in many places directly :) > At this point it's rule (3) that I am still struggling with a bit: > > 50 3) If the code attempts to gain a reference to a kref-ed structure > 51 without already holding a valid pointer, it must serialize access > 52 where a kref_put() cannot occur during the kref_get(), and the > 53 structure must remain valid during the kref_get(). > > In this example, every call to put_entry() results in locking and > unlocking the mutex. But if I am following this right, that is only > because the entry at the head of the list is removed from the list > when and only when the last reference to it is released. It has nothing to do with a list, don't get hung up on a list with a kref, it's not needed, just used as an example here. > If the list_del() happened for some other cause (say a timer expired > or user space sent a message to delete the entry), then the taking of > the mutex in put_entry() wouldn't be necessary, right? No, it still would be. > For instance, this would be valid, wouldn't it? > > static void release_entry(struct kref *ref) > { > struct my_data *entry = container_of(ref, struct my_data, refcount); > > kfree(entry); > } > > static void put_entry(struct my_data *entry) > { > kref_put(&entry->refcount, release_entry); > } > > static void del_entry(struct my_data *entry) > { > mutex_lock(&mutex); > list_del(&entry->link); > mutex_unlock(&mutex); > put_entry(entry); > } Nope. Don't get list entries confused with an object that a kref happens to maintain. You still need to lock the kref itself, two different people could be calling put_entry() at the same time, right? > In this example, threads that want access to an entry do need to take > the mutex in order to get it from the list in get_entry(), but when > they are done with it but don't want it deleted from the list they can > call put_entry() and no mutex need be taken. The only time the mutex > need be taken is when the caller wants to also delete the entry from > the list, which is what del_entry() is for. > > Put another way, the mutex is really there to serialize access to the > list, right? Again, don't focus on the list, and you should be fine. Do you have a real-world use of kref that you are curious about? That might help out more here. thanks, greg k-h _______________________________________________ Kernelnewbies mailing list Kernelnewbies@xxxxxxxxxxxxxxxxx http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
