question about kref API
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. 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. 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, so it is unclear to me what the value of the reference counts in these my_data structures are at the time they are enqueued to the list. Put another way, it's not clear to me from reading this whether kref_init(&(entry->refcount)) was called before the instances of struct my_data were put into the list. So I have two interpretations of what is being illustrated with this sample code and neither makes much sense to me. 1) The krefs are initialized before they go into the list. If the krefs in the instances of struct my_data are initialized by kref_init() before they go into the list and thus start off with a 1, then it would seem to me that the mutex would not be necessary so long as the number of calls to put_entry() never exceeds the number of calls to get_entry(). However in this case, (so long as the number of calls to put_entry() never exceeds the number of calls to get_entry()), the release function would never get called and the entries would remain in the list because the reference count would never get down to 0. So this doesn't seem like a correct interpretation to me unless the assumption is that what is being illustrated is a case where the number of calls to put_entry() equals the number of calls to get_entry() plus 1, the final call to put_entry() being the one that causes the deletion of the entry from the list and its freeing, in which case the mutex is clearly necessary. But this usage (# of puts == # of gets + 1) seems somewhat artificial and contrary to my initial assumption. 2) The krefs are not initialized before they go into the list. On the other hand, if the krefs are not initialized by kref_init() before they go into the list and thus contain 0, then the first call to kref_get() to retrieve a given entry at line 160 would seem problematic since in recent kernels (going back at least 4 years near as I can tell), doing that would generate a warning: 41 static inline void kref_get(struct kref *kref) 42 { 43 /* If refcount was 0 before incrementing then we have a race 44 * condition when this kref is freeing by some other thread right now. 45 * In this case one should use kref_get_unless_zero() 46 */ 47 WARN_ON_ONCE(atomic_inc_return(&kref->refcount) < 2); 48 } It doesn't seem reasonable to me that sample code like this would be written such that it generated warnings when executed. So this doesn't seem like a valid interpretation either. If someone knowledgeable of this interface could explain whether in this sample code the krefs are initialized via kref_init() before they are put into the list, I think it would help a lot in my attempt to understand it. Thanks, Jeff Haran
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.
so it is unclear to me what the value of the reference counts in these my_data structures are at the time they are enqueued to the list. Put another way, it's not clear to me from reading this whether kref_init(&(entry->refcount)) was called before the instances of struct my_data were put into the list.
Yes it was.
So I have two interpretations of what is being illustrated with this sample code and neither makes much sense to me.
1) The krefs are initialized before they go into the list.
Yes.
If the krefs in the instances of struct my_data are initialized by kref_init() before they go into the list and thus start off with a 1, then it would seem to me that the mutex would not be necessary so long as the number of calls to put_entry() never exceeds the number of calls to get_entry().
The mutex is needed as multiple threads could be calling kref_put at the same time if you don't have that. Best thing is, don't use the "raw" kref_put() calls, use the kref_put_mutex() or kref_put_spinlock_irqsave() call instead. It's much easier that way and arguably, I should have done that when I created the API over a decade ago.
2) The krefs are not initialized before they go into the list.
Not true, that will not work. Look at the users of kref in the kernel, they have to be initialized before anyone else can ever "touch" them, it's that simple. What were you looking to use a kref for? Hope this helps, greg k-h
-----Original Message----- From: Greg KH [mailto:greg@kroah.com] Sent: Monday, July 21, 2014 7:18 PM To: Jeff Haran Cc: kernelnewbies@kernelnewbies.org 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.
You are referring to line 147 here, right? That creates an empty list if I am following the code correctly. 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.
so it is unclear to me what the value of the reference counts in these my_data structures are at the time they are enqueued to the list. Put another way, it's not clear to me from reading this whether kref_init(&(entry->refcount)) was called before the instances of struct my_data were put into the list.
Yes it was.
So I have two interpretations of what is being illustrated with this sample code and neither makes much sense to me.
1) The krefs are initialized before they go into the list.
Yes.
If the krefs in the instances of struct my_data are initialized by kref_init() before they go into the list and thus start off with a 1, then it would seem to me that the mutex would not be necessary so long as the number of calls to put_entry() never exceeds the number of calls to get_entry().
The mutex is needed as multiple threads could be calling kref_put at the same time if you don't have that.
Best thing is, don't use the "raw" kref_put() calls, use the kref_put_mutex() or kref_put_spinlock_irqsave() call instead. It's much easier that way and arguably, I should have done that when I created the API over a decade ago.
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. 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? 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); } 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? Or perhaps I am still not understanding rule (3). My apologies if I am coming across as thick here. Just trying to make sure I understand the implications of rule (3). Thanks again, Jeff Haran
On Tue, Jul 22, 2014 at 05:25:03PM +0000, Jeff Haran wrote:
-----Original Message----- From: Greg KH [mailto:greg@kroah.com] Sent: Monday, July 21, 2014 7:18 PM To: Jeff Haran Cc: kernelnewbies@kernelnewbies.org 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
-----Original Message----- From: Greg KH [mailto:greg@kroah.com] Sent: Tuesday, July 22, 2014 4:47 PM To: Jeff Haran Cc: kernelnewbies@kernelnewbies.org Subject: Re: question about kref API
On Tue, Jul 22, 2014 at 05:25:03PM +0000, Jeff Haran wrote:
-----Original Message----- From: Greg KH [mailto:greg@kroah.com] Sent: Monday, July 21, 2014 7:18 PM To: Jeff Haran Cc: kernelnewbies@kernelnewbies.org 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?
Ok, so that's an issue I don't understand. Looking at what kref_put() does, I don't see why any external locking would be required to serialize concurrent callers to put_entry(). As I am sure you are aware, kref_put() ends up being a call to this, where the count passed in is 1: 68 static inline int kref_sub(struct kref *kref, unsigned int count, 69 void (*release)(struct kref *kref)) 70 { 71 WARN_ON(release == NULL); 72 73 if (atomic_sub_and_test((int) count, &kref->refcount)) { 74 release(kref); 75 return 1; 76 } 77 return 0; 78 } atomic_sub_and_test() is in of itself thread safe, at least that's the way I read its header comment block. In this case the refcount field is atomically decremented by 1 and if the result of that decrementation is 0, it returns TRUE otherwise FALSE. That means that if there are N concurrent callers to kref_put() when refcount initially contains N, atomic_sub_and_test() will return TRUE for 1 and only 1 of those callers. All of the other callers will see atomic_sub_and_test() return FALSE and simply return 0. That means the release function can get called only once. And since that will only happen when refcount has gotten down to 0, there will be no other existing references to the kref as all the other callers must have already proceeded far enough so that their calls to atomic_sub_and_test() have completed. They have nothing left to do but return 0 to their callers and if those callers are following the rules they will make no further references to the entry. So I don't see how any sort of race condition is possible among concurrent callers to put_entry() here. There also doesn't seem to be any sort of race possible between concurrent callers to put_entry() and get_entry(). If put_entry() finds that the reference count has gotten down to zero as part of its call to atomic_sub_and_test(), then the entry can't be in the list so a concurrent caller to get_entry() will either see that the list empty and never take a new reference or it will take a reference to the next entry in the list. My apologies for seeming to rat hole on this, but I must be missing something fundamental here since I really don't see any race conditions for your proposed put side locking to prevent. The usage of the atomic refcount seem to be sufficient in this case. That's the beauty of the API, or at least that's what I thought was the beauty of it. :) Clearly there are potential performance benefits in not needing to take locks or mutexes when they are not necessary. If you could elaborate on where the race condition is here, I think you'd being doing both me and the community a great service. Thanks again, Jeff Haran
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
-----Original Message----- From: kernelnewbies-bounces@kernelnewbies.org [mailto:kernelnewbies- bounces@kernelnewbies.org] On Behalf Of Jeff Haran Sent: Wednesday, July 23, 2014 10:33 AM To: 'Greg KH' Cc: kernelnewbies@kernelnewbies.org Subject: RE: question about kref API
-----Original Message----- From: Greg KH [mailto:greg@kroah.com] Sent: Tuesday, July 22, 2014 4:47 PM To: Jeff Haran Cc: kernelnewbies@kernelnewbies.org Subject: Re: question about kref API
On Tue, Jul 22, 2014 at 05:25:03PM +0000, Jeff Haran wrote:
-----Original Message----- From: Greg KH [mailto:greg@kroah.com] Sent: Monday, July 21, 2014 7:18 PM To: Jeff Haran Cc: kernelnewbies@kernelnewbies.org 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...]
Apologies on the long lines of my most recent post. While composing it, I terminated each line with a return at short line lengths. But then Outlook crams them all back together again when I hit send. I'll do better next time.
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?
Ok, so that's an issue I don't understand. Looking at what kref_put() does, I don't see why any external locking would be required to serialize concurrent callers to put_entry(). As I am sure you are aware, kref_put() ends up being a call to this, where the count passed in is 1:
68 static inline int kref_sub(struct kref *kref, unsigned int count, 69 void (*release)(struct kref *kref)) 70 { 71 WARN_ON(release == NULL); 72 73 if (atomic_sub_and_test((int) count, &kref->refcount)) { 74 release(kref); 75 return 1; 76 } 77 return 0; 78 }
atomic_sub_and_test() is in of itself thread safe, at least that's the way I read its header comment block. In this case the refcount field is atomically decremented by 1 and if the result of that decrementation is 0, it returns TRUE otherwise FALSE. That means that if there are N concurrent callers to kref_put() when refcount initially contains N, atomic_sub_and_test() will return TRUE for 1 and only 1 of those callers. All of the other callers will see atomic_sub_and_test() return FALSE and simply return 0. That means the release function can get called only once. And since that will only happen when refcount has gotten down to 0, there will be no other existing references to the kref as all the other callers must have already proceeded far enough so that their calls to atomic_sub_and_test() have completed. They have nothing left to do but return 0 to their callers and if those callers are following the rules they will make no further references to the entry. So I don't see how any sort of race condition is possible among concurrent callers to put_entry() here.
There also doesn't seem to be any sort of race possible between concurrent callers to put_entry() and get_entry(). If put_entry() finds that the reference count has gotten down to zero as part of its call to atomic_sub_and_test(), then the entry can't be in the list so a concurrent caller to get_entry() will either see that the list empty and never take a new reference or it will take a reference to the next entry in the list.
My apologies for seeming to rat hole on this, but I must be missing something fundamental here since I really don't see any race conditions for your proposed put side locking to prevent. The usage of the atomic refcount seem to be sufficient in this case. That's the beauty of the API, or at least that's what I thought was the beauty of it. :)
Clearly there are potential performance benefits in not needing to take locks or mutexes when they are not necessary. If you could elaborate on where the race condition is here, I think you'd being doing both me and the community a great service.
Thanks again,
Jeff Haran
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
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On Wed, Jul 23, 2014 at 05:33:19PM +0000, Jeff Haran wrote:
Clearly there are potential performance benefits in not needing to take locks or mutexes when they are not necessary.
Of course there are. But trust me, you need to use a lock here, as the document tries to explain, otherwise your code is broken.
If you could elaborate on where the race condition is here, I think you'd being doing both me and the community a great service.
Nice try with the "Do it for the community because I don't understand this!" appeal, that doesn't work for me, sorry... I think you need to go look at the code closer and not get confused with functions like kref_sub(), which you should never use unless you know exactly what you are doing, and even then, I strongly discourage their use (which is why there are only 2 users in the kernel.) Focus on the "normal" kref functions, and again, look at in-kernel users for examples of how to use this properly. Also, what are you trying to gain here, do you want to use a kref in your code? If so, please submit patches showing your usage and I will be glad to review them. greg k-h
-----Original Message----- From: 'Greg KH' [mailto:greg@kroah.com] Sent: Wednesday, July 23, 2014 10:48 AM To: Jeff Haran Cc: kernelnewbies@kernelnewbies.org Subject: Re: question about kref API
On Wed, Jul 23, 2014 at 05:33:19PM +0000, Jeff Haran wrote:
Clearly there are potential performance benefits in not needing to take locks or mutexes when they are not necessary.
Of course there are. But trust me, you need to use a lock here, as the document tries to explain, otherwise your code is broken.
If you could elaborate on where the race condition is here, I think you'd being doing both me and the community a great service.
Nice try with the "Do it for the community because I don't understand this!" appeal, that doesn't work for me, sorry...
Boy, I try to be as deferential as possible here and you come back with this snarky response. My experience is when software developers can't or won't explain how their code works, it's because they don't understand it themselves. I won't bother you again. Thanks for your time.
I think you need to go look at the code closer and not get confused with functions like kref_sub(), which you should never use unless you know exactly what you are doing, and even then, I strongly discourage their use (which is why there are only 2 users in the kernel.) Focus on the "normal" kref functions, and again, look at in-kernel users for examples of how to use this properly.
Also, what are you trying to gain here, do you want to use a kref in your code? If so, please submit patches showing your usage and I will be glad to review them.
greg k-h
On Wed, Jul 23, 2014 at 05:55:50PM +0000, Jeff Haran wrote:
-----Original Message----- From: 'Greg KH' [mailto:greg@kroah.com] Sent: Wednesday, July 23, 2014 10:48 AM To: Jeff Haran Cc: kernelnewbies@kernelnewbies.org Subject: Re: question about kref API
On Wed, Jul 23, 2014 at 05:33:19PM +0000, Jeff Haran wrote:
Clearly there are potential performance benefits in not needing to take locks or mutexes when they are not necessary.
Of course there are. But trust me, you need to use a lock here, as the document tries to explain, otherwise your code is broken.
If you could elaborate on where the race condition is here, I think you'd being doing both me and the community a great service.
Nice try with the "Do it for the community because I don't understand this!" appeal, that doesn't work for me, sorry...
Boy, I try to be as deferential as possible here and you come back with this snarky response.
Snarky? You were asking me to take time and personally walk you through the kref code and your misconceptions of it. I've written this document, given numerous presentations on it, wrote a published conference paper on the topic, as well as maintining the kernel code for over a decade now. Trying to say "do it for the community" implies my previous work was somehow insufficient and I should do more to appease your specific need. And that need is still unknown to me, you never said _why_ you wanted to understand the kref code. Which is important to me.
My experience is when software developers can't or won't explain how their code works, it's because they don't understand it themselves.
Yeah, that's it, really, I don't understand it at all, talk about snark :) I asked you to read the in-kernel usage of the code, instead of this contrived example code, which you didn't do. I asked you to provide a use-case or example of where you wanted to use this code, which you didn't do. Instead you insist that I educate you without doing much work on your own, which is not how a community works, sorry. Best of luck in your education process. greg k-h
On Wed, 23 Jul 2014 17:55:50 -0000, Jeff Haran said:
From: 'Greg KH' [mailto:greg@kroah.com]
If you could elaborate on where the race condition is here, I think you'd being doing both me and the community a great service.
Nice try with the "Do it for the community because I don't understand this!" appeal, that doesn't work for me, sorry...
Boy, I try to be as deferential as possible here and you come back with this snarky response.
My experience is when software developers can't or won't explain how their code works, it's because they don't understand it themselves.
On the other hand, the kernel community does expect you to do a large chunk of the learning for yourself - first, because Greg KH isn't a scalable resource, and second, because if you take the time to figure it out for yourself, you'll understand it better than if Greg walked you through it.
participants (4)
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'Greg KH' -
Greg KH
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