A query regarding synchronised functions, using shared object

Question

Hi all.
I have this little query, regarding the functions that are synchronised, based on accessing the lock to the object, which is being used for synchronizing.
Ok, I will clear myself with the following example :
class First
int a;
static int b;
public void func_one()
synchronized((Integer) a)
{ // function logic
} // End of func_one
public void func_two()
synchronized((Integer) b)
{ / function logic
} // End of func_two
public static void func_three()
synchronized((Integer) a)
{ // function logic
} // End of func_three, WHICH IS ACTUALLY NOT ALLOWED,
// just written here for completeness.
public static void func_four()
synchronized((Integer) b)
{ / function logic
} // End of func_four
First obj1 = new First();
First obj2 = new First();
Note that the four functions are different on the following criteria :
a) Whether the function is static or non-static.
b) Whether the object on which synchronization is based is a static, or a non-static member of the class.
Now, first my-thoughts; kindly correct me if I am wrong :
a) In case 1, we have a non-static function, synchronized on a non-static object. Thus, effectively, there is no-synchronisation, since in case obj1 and obj2 happen to call the func_one at the same time, obj1 will obtain lock for obj1.a; and obj2 will obtain lock to obj2.a; and both can go inside the supposed-to-be-synchronized-function-but-actually-is-not merrily.
Kindly correct me I am wrong anywhere in the above.
b) In case 2, we have a non-static function, synchronized on a static object. Here, again if obj1, and obj2 happen to call the function at the same time, obj1 will try to obtain lock for obj1.a; while obj2 will try to obtain lock for obj2.a. However, since obj1.a and obj2.a are the same, thus we will indeed obtain sychronisation.
Kindly correct me I am wrong anywhere in the above.
c) In case 3, we have a static function , synchronized on a non-static object. However, Java does not allow functions of this type, so we may safely move forward.
d) In case 4, we have a static function, synchronized on a static object.
Here, again if obj1, and obj2 happen to call the function at the same time, obj1 will try to obtain lock for obj1.a; while obj2 will try to obtain lock for obj2.a. However, since obj1.a and obj2.a are the same, thus we will indeed obtain sychronisation. But we are only partly done for this case.
First, Kindly correct me I am wrong anywhere in the above.
Now, I have a query : what happens if the call is made in a classically static manner, i.e. using the statement "First.func_four;".
Another query : so far we have been assuming that the only objects contending for the synchronized function are obj1, and obj2, in a single thread. Now, consider this, suppose we have the same reference obj1, in two threads, and the call "obj1.func_four;" happens to occur at the same time from each of these threads. Thus, we have obj1 rying to obtain lock for obj1.a; and again obj1 trying to obtain lock for obj1.a, which are the same locks. So, if obj1.a of the first thread obtains the lock, then it will enter the function no-doubt, but the call from the second thread will also succeed. Thus, effectively, our synchronisation is broken.
Or am I being dumb ?
Looking forward to replies..
Ashutosh

Answer 1

a) In case 1, we have a non-static function, synchronized on a non-static object. Thus, effectively, there is no-synchronisationThere is no synchronization between distinct First objects, but that's what you specified. Apart from the coding bug noted below, there would be synchronization between different threads using the same instance of First.
b) In case 2, we have a non-static function, synchronized on a static object. Here, again if obj1, and obj2 happen to call the function at the same time, obj1 will try to obtain lock for obj1.a; while obj2 will try to obtain lock for obj2.a.obj1/2 don't call methods or try to obtain locks. The two different threads do that. And you mean First.b, not obj1.b and obj2.b, but see also below.
d) In case 4, we have a static function, synchronized on a static object. Here, again if obj1, and obj2 happen to call the function at the same time, obj1 will try to obtain lock for obj1.a; while obj2 will try to obtain lock for obj2.a.Again, obj1/2 don't call methods or try to obtain locks. The two different threads do that. And again, you mean First.b. obj1.b and obj2.b are the same as First.b. Does that make it clearer?
Now, I have a query : what happens if the call is made in a classically static manner, i.e. using the statement "First.func_four;".That's what happens in any case whether you write obj1.func_four(), obj2.func)four(), or First.func_four(). All these are identical when func_four(0 is static.
Now, consider this, suppose we have the same reference obj1, in two threads, and the call "obj1.func_four;" happens to occur at the same time from each of these threads. Thus, we have obj1 rying to obtain lock for obj1.aNo we don't, we have a thread trying to obtain the lock on First.b.
and again obj1 trying to obtain lock for obj1.aYou mean obj2 and First.b, but obj2 doesn't obtain the lock, the thread does.
which are the same locks. So, if obj1.a of the first thread obtains the lock, then it will enter the function no-doubt, but the call from the second thread will also succeed.Of course it won't. Your reasoning here makes zero sense..Once First.b is locked it is locked. End of story.
Thus, effectively, our synchronisation is broken.No it isn't. The second thread will wait on the same First.b object that the first thread has locked.
However in any case you have a much bigger problem here. You're autoboxing your local 'int' variable to a possibly brand-new Integer object every call, so there may be no synchronization at all.
You need:
Object a = new Object();
static Object b = new Object();