Java and exact reference size for objects, array and primitive types

I would like to know exactly the real space allocated in memory for an object.

I try to explain with some example: using a 64 bit JVM, pointer size should be 8 bytes, so:

Object singletest = new Object(); will take 8 bytes to reference the Object plus the size of the Object
Object arraytest = new Object[10]; will take 8 byte to reference the position where the array is stored plus 8*10 bytes to store the array plus the size of each Object
int singleint = new int; will take just 2 bytes, because int is a primitive type
int[] arrayint = new int[10]; will take 8 bytes to reference the position and 10*2 bytes for the elements

Moreover, this is the reason why Java allows to write code like this:

int[][] doublearray = new int[2][];
int[0][] = new int[5];
int[1][] = new int[10];

What really happen is that an array will produce a reference (aka pointer) like an object, so it doesn't really matter the size of the second dimension at declaration time (and dimensions can be different, there is no link between them). Then the space taken will be: a reference to doublearray (8 bytes), first dimension is simply a reference to the second one, so other 8 bytes * 2 (first dimension size), and finally 2 bytes * 5 plus 2 bytes * 10.

So, finally, if have a real class like this:

class Test {
   int a, b;
   int getA() {return A};
   void setA(int a) {this.a = a;}
   int getB() {return B};
   void setB(int b) {this.b = b;}
}

when I call a new to instantiate, a pointer (or name it reference, because it's Java) of 8 bytes will be used plus 2+2bytes to store the integers into the class.

The questions are: am I right or I wrote total nonsense? Moreover, when I don't instantiate an object but I just declare it, 8 bytes will be allocated for further use or not? And what if I assign a null value?

Meanwhile for primitive type I'm quite sure that just declaring it will allocate the requested space (if I declare an "int i" then I can immediately call i++ because no reference are used, just a portion of memory is setted on "0").

I searched on internet without clever response... I know that I wrote a lot of questions, but any help will be appreciated! (and maybe I'm not the only one interested)

using a 64 bit JVM, pointer size should be 8 bytes,

Actually its usually 32-bit unless you have a maximum heap size of 32 GB or more heap. This is because Java uses references, not pointers (and each object is on an 8 byte, not 1 boundary)

The JVM can change the size of a reference depending on which JVM you use and what the maximum heap size is.

Object singletest = new Object(); will take 8 bytes to reference the Object plus the size of the Object

The object will use about 16 bytes of heap. It may or may not use 4 bytes of stack, but it could just use a register.

Object arraytest = new Object[10];

This will use about 16 bytes for the header, plus 10 times the reference sizes (about 56 bytes in total)

int singleint = new int; will take just 2 bytes, because int is a primitive type

int is always 32-bit bit, you can't create a newprimitive. As its notionally on the stack it might use 4-bytes of stack or it might only use a register.

int[] arrayint = new int[10]; will take 8 bytes to reference the position and 10*2 bytes for the elements

Again the object is likely to be the same size as the new Object[10] (56 bytes)

int[][] doublearray = new int[2][];
int[0][] = new int[5];
int[1][] = new int[10];

I wouldn't call it a doublearray as it could be confused with double[]

However the size is likkely to be about 16 + 2 * 4 for doublearray and 16 + 5*4 + 4 (for padding) and 16 + 10 * 4.

Memory allocated on the heap is aligned to an 8 byte boundary.

I call a new to instantiate, a pointer (or name it reference, because it's Java) of 8 bytes will be used plus 2+2bytes to store the integers into the class.

The reference is on the stack and usually this is not included.

The object has a header of about 16 bytes and the int values take up 2 * 4 bytes.