How does synchronized work in Java
First, here's a sample:
public class Deadlock {
static class Friend {
private final String name;
public Friend(String name) {
this.name = name;
}
public String getName() {
return this.name;
}
public synchronized void bow(Friend bower) {
System.out.format("%s: %s has bowed to me!%n",
this.name, bower.getName());
bower.bowBack(this);
}
public synchronized void bowBack(Friend bower) {
System.out.format("%s: %s has bowed back to me!%n",
this.name, bower.getName());
}
}
public static void main(String[] args) {
final Friend alphonse = new Friend("Alphonse");
final Friend gaston = new Friend("Gaston");
new Thread(new Runnable() {
public void run() { alphonse.bow(gaston); }
}).start();
new Thread(new Runnable() {
public void run() { gaston.bow(alphonse); }
}).start();
}
}
What I don't get is how the blockage occurs. The main function initiates two threads that each begin their own bows.
What exactly does 'synchronized' block? The same function running for the same object (as I originally thought)? The same function for all objects of the same class? All synchronized functions for the same object? All synchronized functions for all objects of the same class?
Help me out here.
In Java, each Object provides the ability for a Thread to synchronize , or lock, on it. When a method is synchronized, the method uses its object instance as the lock. In your example, the methods bow and bowBack are both synchronized , and both are in the same class Friend . This means that any Thread executing these methods will synchronize on a Friend instance as its lock.
A sequence of events which will cause a deadlock is:
alphonse.bow(gaston) , which is synchronized on the alphonse Friend object. This means the Thread must acquire the lock from this object. gaston.bow(alphonse) , which is synchronized on the gaston Friend object. This means the Thread must acquire the lock from this object. bowback and waits for the lock on gaston to be released. bowback and waits for the lock on alphonse to be released. To show the sequence of events in much more detail:
main() begins to execute in the main Therad (call it Thread #1), creating two Friend instances. So far, so good. new Thread(new Runnable() { ... . Thread #2 calls alphonse.bow(gaston) , which is synchronized on the alphonse Friend object. Thread #2 thus acquires the "lock" for the alphonse object and enters the bow method. gaston.bow(alphonse) , which is synchronized on the gaston Friend object. Since no-one has yet acquired the "lock" for the gaston object instance, Thread #3 successfully acquires this lock and enters the bow method. bower.bowBack(this); with bower being a reference to the instance for gaston . This is the logical equivalent of a call of gaston.bowBack(alphonse) . Thus, this method is synchronized on the gaston instance. The lock for this object has already been acquired and is held by another Thread (Thread #3). Thus, Thread #2 has to wait for the lock on gaston to be released. The Thread is put into a waiting state, allowing Thread #3 to execute further. bowback , which in this instance is logically the same as the call alphonse.bowBack(gaston) . To do this, it needs to acquire the lock for the alphonse instance, but this lock is held by Thread #2. This Thread is now put into a waiting state. And you are now in a position where neither Thread can execute. Both Thread #2 and Thread #3 are waiting for a lock to be released. But neither lock can be released without a Thread making progress. But neither thread can make progress without a lock being released.
Thus: Deadlock!
Deadlocks very often depend on a specific sequence of events occurring, which can make then difficult to debug since they can be difficult to reproduce.
Synchronized has two effects:
So in short, it blocks any invocations of synchronised methods on the same object.
All synchronized functions for the same object. Marking a method "synchronized" is very similar to putting a "synchronized (this) {" block around the entire contents of the method. The reason I don't say "identical" is because I don't know offhand whether the compiler emits the same bytecode or not, but AFAIK the defined runtime effect is the same.
The deadlock is a classic locking inversion. One thread locks alphonse. Then (or simultaneously on a multi-core system) the other thread locks gaston. This part requires that the scheduling of the threads just so happens to interleave at the right points.
Each thread (in whatever order or simultaneously) then attempts to acquire a lock which is already held by the other thread, and hence each thread goes to sleep. Neither will wake until the other releases its lock, but neither will release its lock until it wakes (or is terminated).
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