reference type size in java

Question

Why does a reference type in java take 8 bytes? Why not less or more than 8 bytes?

Myer

Answer 1

Actually, it is nowhere specified how much bytes a reference variable shall have, and in fact, it is not everywhere the same.

Common virtual machines for 32-bit systems (i.e. systems with 32 bit adresses in the memory bus) usually use 32-bit (= 4 bytes, same as int and float) as the size for object references, while virtual machines for 64-bit systems often use the native address size 64 bits (= 8 bytes) for these. (Note that most 64 bit systems can run 32-bit programs, too, so often even there you'll be using a 32 bit VM.)

It is simply a matter of simplifying the implementation, if you can use an actual memory address for your references instead of something else.

As this increases the size of memory used (and often we don't actually need to access that much memory), from Java 7 on the 64-bit HotSpot VM will be able to use 32-bit references under certain conditions, i.e. when the heap is smaller than 32 GB (8·2³² bytes). To convert them to a actual memory address, they are multiplied by 8 (as objects will be aligned on 8-byte-boundaries) and then added to the base address (if this is not zero). (For heaps smaller than 4 GB, we don't need the multiplication step.)

Other VMs might use similar tricks.

Nedry

64 bit JVMs don't necessarily use 64 bit references. Hotspot for example uses 32 bits by default. – Syllabub

Thanks for the correction ... do you have a link to the relevant documentation? – Zeitler

See also: docs.oracle.com/javase/7/docs/technotes/guides/vm/… – Syllabub

@Syllabub Thanks, I added this to my answer. – Zeitler

Answer 2

This is because a reference variable doesn't actually hold the object. It's a way to reach the object. The way JVM manages this is something of our least concern. You can think of it as an address to the location of the object in the heap. But it need not be as straight forward as an address.

Expurgatory

In fact it probably can't, as the GC moves objects around in the physical memory a lot, so the JVM needs an extra level of indirection to keep references up to date. – Yahairayahata

Answer 3

Let me give an example to help you understand better

    String myName = new String("John");
    String yourName = new String("John");
    if (myName == yourName) 
    {
        System.out.println("Refereces point to Same objects on heap");
    } else 
    {
        System.out.println("Refereces point to different objects on heap");
    }

and output is Refereces point to different objects on heap.

Here myName and yourName are two references which point to objects of type String(memory allocated on Heap). Note memory for reference variables is allocated on stack and not on the Heap. Both the reference variables merely have the address(or similar unique abstraction) of the String object which resides on the Heap. The size of address will be constant for a particular architecture of OS. In case of 32 bits it will be 4 bytes where as for 64 bits it will be 8 bytes.

So size of objects pointed be the reverence variable may change but the size of references remain the same as they simply carry address which remains constant for any particular architecture.

Schuler

Answer 4

8 Bytes is the size of a Long variable in Java and allows for 1.01457092 × 10^19 references to be referenced, that should be enough for most applications, and bigger than most hard drives can address considering that each of those reference points needs to also hold data...

Edit: Having read one of the other answers I wanted to clarify that the reference is just a pointer to the data and doesn't contain the data, the Java architects decided to use 64 bit values to denote references because it gives you a huge address space.

Tropic

Answer 5

Values of type reference can be thought of as pointers to objects. the 8 byte should adress this in the heap.

Read about The Structure of the Java Virtual Machine

Though

Answer 6

because 64bit is enough for most hardware address space. and the reference can be seen as a pointer.

Williamson