The truth about objects, references, instances and cloning (Trees and Lists

Question

Hello
As a few of you may have know from my postings over the past few weeks, ive been building my first real java ap and trying to learn along the way, hitting a few snags here and there.
One main snag i seem to have been hitting a fair bit is objects and instances, im kinda learning on how they work and what they mean but as a php programmer it seems to be very different.
as pointed out to me if you have objectA=objectB then in php you have two objects that share the same value but in java you have still one object value and one object pointing to the value.
sooo my first question is
if i have this
Object objA=new Object()
Object objB=objA()then object A creates a new memory allocation for the object data right? and object B is simply a reference to the object A memory allocation right?
so therefore there is no instance of objectB? or is there an instance of object B but it just takes up the same memory allocation as object A?
anyway, what is the point of being able to say objB=objA what can that be used for if not for making a copy of an object (i understand now that it doesnt copy an object)
My second question (which i think i actually understand now, but im posting it to see if the answers may help others)
If i have a List structure (e.g. arraylist) then when i add a datatype such as a treemap does it not add the changed treemap, ill try and explain with code
ArrayList mylist=new ArrayList()
Treemap myTree=new Treemap()
myTree.put("hello","howdy");
myTree.put("bye","cya");
mylist.put(myTree);
System.out.println(mylist.toString());
myTree.put("another","one");
mylist.put(myTree);
System.out.println(mylist.toString());now to be honest ive not actually tried that code and it may actually procude the right results, but from my experience so far that similar sort of thing hasnt been working (by the way i know the above code wont compile as is :p)
anyway what i assume is that will output this
[Hello,howdy,bye,cya] <<this is the first system out
[Hello,howdy,bye,cya,another,one  <<this is the second system out
Hello,howdy,bye,cya,another,one] <<this is the second system out
where it should produce this
[Hello,howdy,bye,cya,
Hello,howdy,bye,cya,another,one]
Why when I update the treemap does the arraylist change, is this because the thing that is being put in the array list is infact not an object but simply a reference to an object?
thanks for everyones help so far :)

Answer 1

as pointed out to me if you have objectA=objectB then
in php you have two objects that share the same value
but in java you have still one object value and one
object pointing to the value.Some years ago, I built a small website managing data saved in an XML file using PHP. Being used to Java, I used an OO approach and created classes managing data structures and having setter/getter methods. The fact that PHP did actually create a copy of any object being passed as and argument made me crazy. This way, I wasn't able to reach the same instances from several different places in my code. Instead, I had to put a certain operator ("&" I think) before any reference to force PHP not to make a copy but pass the reference. I also found a note on the PHP site saying that copying was faster than passing the reference, which I actually couldn't believe. If I'm not wrong, they changed the default behaviour in PHP5.
if i have this
Object objA=new Object()
Object objB=objA()then object A creates a new memory allocation for the
object data right? and object B is simply a reference
to the object A memory allocation right?The statement "new <Class>" allocates memory. The reference itself just takes up a few bytes in order to maintain some administrative data (such as the memory address). References are basically like pointers in C, though there's no pointer arithmetics and they can't point to any arbitrary place in memory (only a legal instance or null).
so therefore there is no instance of objectB? or is
there an instance of object B but it just takes up
the same memory allocation as object A?There is an instance of objectB, but it's just the same instance that objectA is pointing at. Both references objectA and objectB contain the same memory address. That's why they'd produce a result of "true" if compared using the "==" operator.
anyway, what is the point of being able to say
objB=objA what can that be used for if not for making
a copy of an object (i understand now that it doesnt
copy an object)Sometimes you don't want to make a copy, as 1) it consumes more memory and 2) you'd lose the reference to it (making it more difficult to tell what instance is actually accessed). If there was no way to pass a reference to an object into a method, methods wouldn't be able to affect the original instance.
Why when I update the treemap does the arraylist
change, is this because the thing that is being put
in the array list is infact not an object but simply
a reference to an object?The ArrayList doesn't change at all. It's the instance that's being changed, and the ArrayList has a reference to it.