Java – Implementing a global lock in Java

globaljavalockingmultithreading

I have a relatively simple (perhaps stupid) question regarding synchronisation in Java.

I have synchronisation blocks that acquire locks on various objects throughout my code. In some scenarios, I want to acquire a global lock that subsumes every other synchronisation statement in my code.

Is there a fancy way to do this in Java without re-writing all the current synchronisation code?

For example,

Thread t1

synchronized (o1)
{
    synchronized (o2)
    {
        // ...
    }
}

Thread t2

synchronized (global_lock)
{
    // ...
}

When thread t2 is inside the synchronised block, thread t1 should not be allowed to acquire the locks on o1 and o2.

Many thanks
if

Best Solution

It's not possible;
It's a really bad idea ( sorry ).

It's deadlock-prone, since it forces you to have a pre-determined locking order for all locks, no matter where they are.

Usually it's a good idea, if you need to acquire two locks, to always have a predetermined order:

synchronized(LOCK1) {
  synchronized(LOCK2) {

  }
}

But a global lock would need some sort of protocol - a global aquisition order - for all the locks. And that may not be possible at all. Most locks guard specific, self-contained, critical sections. They would be unaware that someone would 'yank' them out and aquire them and therefore would not be written to handle this situation.

So it's not possible, and you should be happy it's not. Although it seems to be the easy way out, it would bring a lot of pain.

Related Solutions

Java – What are the differences between a HashMap and a Hashtable in Java

There are several differences between HashMap and Hashtable in Java:

Hashtable is synchronized, whereas HashMap is not. This makes HashMap better for non-threaded applications, as unsynchronized Objects typically perform better than synchronized ones.
Hashtable does not allow null keys or values. HashMap allows one null key and any number of null values.
One of HashMap's subclasses is LinkedHashMap, so in the event that you'd want predictable iteration order (which is insertion order by default), you could easily swap out the HashMap for a LinkedHashMap. This wouldn't be as easy if you were using Hashtable.

Since synchronization is not an issue for you, I'd recommend HashMap. If synchronization becomes an issue, you may also look at ConcurrentHashMap.

Java – Is Java “pass-by-reference” or “pass-by-value”

Java is always pass-by-value. Unfortunately, when we deal with objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.

It goes like this:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    // we pass the object to foo
    foo(aDog);
    // aDog variable is still pointing to the "Max" dog when foo(...) returns
    aDog.getName().equals("Max"); // true
    aDog.getName().equals("Fifi"); // false
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // change d inside of foo() to point to a new Dog instance "Fifi"
    d = new Dog("Fifi");
    d.getName().equals("Fifi"); // true
}

In the example above aDog.getName() will still return "Max". The value aDog within main is not changed in the function foo with the Dog "Fifi" as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.

Likewise:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    foo(aDog);
    // when foo(...) returns, the name of the dog has been changed to "Fifi"
    aDog.getName().equals("Fifi"); // true
    // but it is still the same dog:
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // this changes the name of d to be "Fifi"
    d.setName("Fifi");
}

In the above example, Fifi is the dog's name after call to foo(aDog) because the object's name was set inside of foo(...). Any operations that foo performs on d are such that, for all practical purposes, they are performed on aDog, but it is not possible to change the value of the variable aDog itself.

For more information on pass by reference and pass by value, consult the following SO answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.