How do hashCode() and identityHashCode() work at the back end?

Question

How do Object.hashCode() and System.identityHashCode() work at the back end? Does identityHashCode() return the reference of the object? Does hashCode() depend on the content or address of the object?

What is the difference between hashCode() and identityHashCode()?

Answer 1

How do Object.hashCode() and System.identityHashCode() work at the back end?

Assuming that it hasn't been overridden, the Object.hashCode() method simply calls System.identityHashCode(this).

The exact behavior of System.identityHashCode(Object) depends on the JVM implementation. (The actual implementation on recent Hotspot JVMs is rather clever, but I digress.)

Does identityHashCode() return the reference of the object?

No. It returns an int, and an int cannot hold a reference.

That integer returned by identityHashCode may be related to the (a) machine address for the object, or it may not be¹. The value returned by identityHashCode() is guaranteed not to change for the lifetime of the object. This means that if the GC relocates an object (after an identityHashCode() call) then it cannot use the new object address as the identity hashcode.

Does hashCode() depend on the ? of the object ? == operator how to work in back end.

This doesn't make sense. There is no ? == or ?== operator in Java.

What is the difference between hashCode() and identityHashCode()?

This is partly explained above. Other differences include:

• The hashcode() method is a non-final instance method, and should be overridden in any class where the equals(Object) is overridden. By contrast, identityHashCode(Object) is a static method and therefore cannot be overridden.

• The identityHashCode(Object) method gives you a identifier for an object which can (in theory) be used for other things than hashing and hash tables. (Unfortunately, it is not a unique identifier, but it is guaranteed to never change for the lifetime of the object.)

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^{1 - For current generation JVMs, it is not related to the memory address at all. See @bestsss's answer.}