Why does notifyAll() raise IllegalMonitorStateException when synchronized on Integer?

问题

Why does this test program result in a java.lang.IllegalMonitorStateException?

public class test {
    static Integer foo = new Integer(1);
    public static void main(String[] args) {
        synchronized(foo) {
            foo++;
            foo.notifyAll();
        }
        System.err.println("Success");
    }
}

Result:

Exception in thread "main" java.lang.IllegalMonitorStateException
        at java.lang.Object.notifyAll(Native Method)
        at test.main(test.java:6)

回答1:

You have noted correctly that notifyAll must be called from a synchronized block.

However, in your case, because of auto-boxing, the object you synchronized on is not the same instance that you invoked notifyAll on. In fact, the new, incremented foo instance is still confined to the stack, and no other threads could possibly be blocked on a wait call.

You could implement your own, mutable counter on which synchronization is performed. Depending on your application, you might also find that AtomicInteger meets your needs.

回答2:

You should also be leery of locking or notifying on objects like String and Integer that can be interned by the JVM (to prevent creating a lot of objects that represent the integer 1 or the string "").

回答3:

Incrementing the Integer makes the old foo disappear and be replaced with a brand new object foo which is not synchronized with the previous foo variable.

Here is an implementation of AtomicInteger that erickson suggested above. In this example foo.notifyAll(); does not produce a java.lang.IllegalMonitorStateException beause the AtomicInteger Object is not refreshed when foo.incrementAndGet(); is run.

import java.util.concurrent.atomic.AtomicInteger;

public class SynchronizeOnAPrimitive {
    static AtomicInteger foo = new AtomicInteger(1);
    public static void main(String[] args) {
        synchronized (foo) {
            foo.incrementAndGet();
            foo.notifyAll();
        }
        System.out.println("foo is: " + foo);
    }
}

Output:

foo is: 2

回答4:

As erickson has noted, the code without the postincrement operator works without error:

static Integer foo = new Integer(1);

public static void main(String[] args) {
    synchronized (foo) {
        foo.notifyAll();
    }
    System.out.println("Success");
}

output:

Success

来源：https://stackoverflow.com/questions/260337/why-does-notifyall-raise-illegalmonitorstateexception-when-synchronized-on-int

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java

synchronization

notify

illegalmonitorstateexcep