If I call the Object.hashcode()
method on some object it returns the internal address of the object (default implementation). Is this address a logical or physical address?
In garbage collection, due to memory compaction objects shifting takes place in the memory. If I call hashcode before and after the GC, will it return the same hashcode (it returns) and if yes then why (because of compaction address may change) ?
The point is that hashcodes can be the same without necessarily guaranteeing that the objects are equal, because the "hashing algorithm" used in the hashCode() method might happen to return the same value for multiple objects.
Simply put, hashCode() returns an integer value, generated by a hashing algorithm. Objects that are equal (according to their equals()) must return the same hash code. Different objects do not need to return different hash codes.
If two objects are equal according to the equals(Object) method, then calling the hashCode method on each of the two objects must produce the same integer result. It is not required that if two objects are unequal according to the equals(java. lang.
The hashCode() is a method of Java Integer Class which determines the hash code for a given Integer. It overrides hashCode in class Object. By default, this method returns a random integer that is unique for each instance.
@erickson is more or less correct. The hashcode returned by java.lang.Object.hashCode()
does not change for the lifetime of the object.
The way this is (typically) implemented is rather clever. When an object is relocated by the garbage collector, its original hashcode has to be stored somewhere in case it is used again. The obvious way to implement this would be to add a 32 bit field to the object header to hold the hashcode. But that would add a 1 word overhead to every object, and would waste space in the most common case ... where an Object's hashCode
method is not called.
The solution is to add two flag bits to the object's flag word, and use them (roughly) as follows. The first flag is set when the hashCode
method is called. A second flag tells the hashCode
method whether to use the object's current address as the hashcode, or to use a stored value. When the GC runs and relocates an object, it tests these flags. If the first flag is set and second one is unset, the GC allocates one extra word at the end of the object and stores the original object location in that word. Then it sets the two flags. From then on, the hashCode
method gets the hashcode value from the word at the end of the object.
In fact, an identityHashCode
implementation has to behave this way to satisfy the following part of the general hashCode contract:
"Whenever it is invoked on the same object more than once during an execution of a Java application, the hashCode method must consistently return the same integer, provided no information used in equals comparisons on the object is modified. This integer need not remain consistent from one execution of an application to another execution of the same application."
A hypothetical implementation of identityHashCode()
that simply returned the current machine address of an object would violate the highlighted part if/when the GC moved the object to a different address. The only way around this would be for the (hypothetical) JVM to guarantee that an object never moves once hashCode
has been called on it. And that would lead to serious and intractable problems with heap fragmentation.
No, the default hash code of an object will not change.
The documentation doesn't say that the hash code is the address, it says that it is based on the address. Consider that hash codes are 32 bits, but there are 64-bit JVMs. Clearly, directly using the address wouldn't always work.
The implementation depends on the JVM, but in the Sun (Oracle) JVM, I believe the hash code is cached the first time it's accessed.
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