I have the following code: Note: I simplified the code as much as possible for readability. If I forgot any critical pieces let me know.
public class User(){
private Relations relations;
public User(){
relations = new Relations(this);
}
public getRelations(){
return relations;
}
}
public class Relations(){
private User user;
public Relations(User user){
this.user = user;
}
public synchronized void setRelation(User user2){
Relations relations2 = user2.getRelations();
synchronized(relations2){
storeRelation(user2);
if(!relations2.hasRelation(user))
relations2.setRelation(user);
}
}
public synchronized boolean hasRelation(User user2){
... // Checks if this relation is present in some kind of collection
}
/*Store this relation, unless it is already present*/
private void storeRelation(User user2){
... // Stores this relation in some kind of collection
}
}
This implementation should make sure that for all Relations x, y with:
x.user = u_x
y.user = u_y
the following invariant holds:
x.hasRelation( u_y ) <=> y.hasRelation( u_x )
I believe that holds for the code stated above?
Note: It does of course not hold during the execution of setRelation(..), but at that moment the locks for both relations involved are held by the executing thread so no other thread can read the hasRelation(..) of one of the relations involved.
Assuming that this holds i believe there is still a potential deadlock-risk. Is that correct? And if it is, how can I solve it? I think i would need to obtain both locks in setRelation(..) atomically somehow.
You are correct on both points: your invariant does hold (assuming that I understand correctly what your method-names mean and so on, and assuming that by
if(!relations.hasRelation(user)) relations2.setRelation(user2);
you meant to writeif(!relations2.hasRelation(user)) relations2.setRelation(user);
), but you do have the risk of a deadlock: if one thread needs to obtain a lock onx
and then ony
, and another thread needs to obtain a lock ony
and then onx
, then there's a risk that each thread will succeed in getting its first lock, and thereby prevent the other from getting its second lock.One solution is to enforce a strict universal ordering for getting locks on
Relations
instances. What you do is, you add a constant integer fieldlockOrder
:and a static integer field
currentLockOrder
:and every time you create a
Relations
instance, you set itslockOrder
to the current value ofcurrentLockOrder
, and increment said:such that every instance of
Relations
will have a distinct, immutable value forlockOrder
. YoursetRelation
method would then obtain locks in the specified order:thereby ensuring that if two threads both need to get locks on both
x
andy
, then either they'll both first get locks onx
, or they'll both first get locks ony
. Either way, no deadlock will occur.Note, by the way, that I changed
setRelation
tostoreRelation
.setRelation
would work, but why add that complexity?Also, there's still one thing I don't get: how come
x.setRelation(u_y)
callsx.storeRelation(u_y)
unconditionally, but callsy.setRelation(u_x)
(ory.storeRelation(u_x)
) only ify
doesn't already have the relationship? It doesn't make sense. It seems like either both checks are needed, or neither check is. (Without seeing the implementation ofRelations.storeRelation(...)
, I can't guess which of those is the case.)