here... 

Explanation:
Magic  

1. ArrayList<PVector> history = new ArrayList();
2. 
   
3. void setup() {
4. 
   
5.   PVector d = new PVector();
6.   history.clear();
7. 
   
8.   d.x=100;
9.   d.y=100;
10.   d.z=0;
11.   history.add(d);
12. 
    
13.   d = new PVector();
14.   d.x=200;
15.   d.y=200;
16.   d.z=0;
17.   history.add(d);
18. 
    
19.   println(history.size());
20. }
21. 
    
22. void draw() {
23.   for (PVector thisPVector : history) {
24.     println (thisPVector);
25.   }
26.   //
27. } // func
    
28. //

or like this


Other people can explain this better than I, but the reason is that you give into the arraylist a mere pointer to the address in the memory where d is. Now you change d (1st entry in history changes not but still points to the old adress with a d with a new content now) and give it another pointer to that same adress (same d).
Now both items in arraylist point to that same address. Bad.

With saying

1. d = new PVector();

you start a new object also named d and it's at another memory position. Thus it works.
Or so.

1. ArrayList history = new ArrayList();
2. 
   
3. PVector d = new PVector();
4. d.x=100;
5. d.y=100;
6. d.z=0;
7. history.add(d);
8. 
   
9. d = new PVector();
10. d.x=200;
11. d.y=200;
12. d.z=0;
13. history.add(d);
14. 
    
15. println(history);

I love you dude....

Best answer...

thanks!!!

Nice!   

Perhaps a former explanation would help you too?  

In your original example, when you use -> history. add (d);
You're adding the value inside variable d.

But what is really stored inside it?
Answer: the memory address plus a type reference of an instance of class PVector!

You do that 2 times. However, since the address memory is the same,
you'll always get what's most recently referenced by that portion of memory all the time!

Unless of course, if you create a new instance and obtain a new memory allocation!