class testClass {
static int counter = 0;
synchronized void bump() {
counter++;
}
static synchronized void staticBump() {
counter++;
}
}[java] static synchronized question
Author
I think you are right. You could call staticBump from bump() (which could then be unsynchronised). Another idiom in Java: you can create a new static object that is private to the class, and synchronise on it:
This would be used for more complex cases (protecting an entire class invariant, possibly across more than one instance). It also reduces potential lock contention, but this generally doesn't do much unless you are in the habit of synchronising over random Class instances from other classes, which isn't great style to begin with.
In this specific instance, you could replace the counter with an AtomicInteger.
class TestClass { private static final Object lock = new Object(); private static int counter = 0; void bump() { synchronized(lock) { counter++; } } static void staticBump() { synchronized(lock) { counter++; } }}This would be used for more complex cases (protecting an entire class invariant, possibly across more than one instance). It also reduces potential lock contention, but this generally doesn't do much unless you are in the habit of synchronising over random Class instances from other classes, which isn't great style to begin with.
In this specific instance, you could replace the counter with an AtomicInteger.
No. The static method will synchronize around the static class object. You would need to do this:
class TestClass { static int counter = 0; void bump() { synchronized(TestClass.class) { counter++; } } static synchronized void staticBump() { counter++; }}This topic is closed to new replies.
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