What Happens to Java Object\'s `hashCode()` When It\'s Not Overridden?

Understanding Hash Code in Java: Default Behavior When Not Overridden

In Java, the hashCode() method plays a crucial role in determining an object's unique identifier. As an inherent method, its behavior can vary depending on whether or not it's overridden by the object's class. Let's explore the default implementation of hashCode() when left untouched.

Default Hash Code Generation for Objects

If the hashCode() method is not overridden, the HotSpot JVM (the widely used implementation of Java) employs a mechanism to generate and store a random number in the object's header. Upon subsequent calls to hashCode(), this random number is simply retrieved. This behavior is designed to provide an unpredictable and unique identifier for objects, without any correlation to their content or location.

HotSpot JVM Configuration for Hash Code

The generation strategy for hash codes can be controlled through the -XX:hashCode=n HotSpot JVM option:

• 0 (default in Java 7): Uses a global random generator, prone to race conditions and contention.
• 5 (default in Java 8): Uses a thread-local xor-shift random generator, eliminating previous issues.
• 1: Mixes object pointer with a random value, providing stability between stop-the-world events (for testing/debugging).
• 2: Always returns 1 (for testing/debugging).
• 3: Uses autoincrementing numbers (for testing/debugging, with potential contention).
• 4: Uses a trimmed object pointer (for testing/debugging).

It's important to note that hash codes generated even with -XX:hashCode=4 may not directly reflect the object's memory address due to possible object relocations. Additionally, unbalanced hash tables may arise if the distribution of object addresses is poor.

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