arr1=arr2;
(filling arr1 with the same values as arr2)?
No.
After that, arr1 and arr2 just hold references to the same array object in memory. So, if you change one, you actually change the data in memory, so the other changes accordingly.

arrayCopy might work, but a double-sized array is just an array of arrays, so you have to make the copy for each array. Ie. you loop on the first dimension, and do an arrayCopy of each entry.

Comparison: yes, you have to use a loop. You know they are different as soon as the first difference is found (or if they don't have the same size!).

About the last question: I guess my first answer answered it as well...

Hello PhiLho,

thanks a lot for the explanation.

And finally regarding call by ref / call by val:
a call by ref

I have a question that is related.

I try to implement a function that works recursive:
http://en.wikipedia.org/wiki/Backtracking_search

I had this working in another programming language.
There it was a call by value.

Thus when returning to the function from a previous higher recursion depth
I need the old value to be in my function. This is possible with a call by value,
but not with a call by ref as we have here in p5.

So in order to implement my little algorithm, I'd like to ask how I can implement a
recursive func with a call by value? Or what are the common work arounds?

Thanks!

Greetings, Chrisir  

What algorithm do you want to implement, exactly?
If you need an old value, you can just pass it as a parameter, no?

hello,

thanks a lot.

I'm working on it.

I want to do this kind of function:

http://en.wikipedia.org/wiki/Backtracking_search#Pseudocode

Greetings, Chrisir

Mmm, I better see where you are going, but that's a very generic algorithm and pseudo-code there...

Anyway, Java always passes data by value to its methods. But in the case of objects, the value is just a reference. The object pointed by that reference can be changed, of course.

Here is a simplistic demonstration:

1. void setup()
2. {
3.   P p = new P();
4.   p.val = 11;
5.   p.x = 111;
6.   println(p);
   
7.   Mv(p);
8.   println(p);
9.   Mx(p);
10.   println(p);
11.   Mx(p);
12.   println(p);
13.   exit();
14. }
15.  
16. class P
17. {
18.   int val;
19.   float x;
20.   public String toString() { return "P[val=" + val + ", x=" + x + "]"; }
21. }
22.  
23. void Mv(P p)
24. {
25.   p.val = 42;
26. }
27.  
28. void Mx(P p)
29. {
30.   p.x *= 0.42;
31. }

Thank you so much, I'll give it a try...

I was able to finish this.

It doesn't work in the browser.

But it can solve two small kinds of the game Solitaire.

http://www.openprocessing.org/sketch/65645

Thanks, PhiLho!

 

update:

It now shows the 1st solution of the full English Board, too.