Help comparing two ArrayList initializations instructions.

Question

I found these two methods to initialize ArrayList on a single row in a 3 years old topic here in this forum... so I open a new thread to ask you what you think is the best method I could use. Thank you!
    private static ArrayList<String> strSearchs = new ArrayList<String>(){{add(" state changed from Full to ");}};
    private static ArrayList<String> strSearchs2 = (ArrayList<String>)Arrays.asList(" state changed from Full to ");PS: note that original suggestions was for "List" objects, so I had to add a cast to "ArrayList" in the second instruction.

Answer 1

Pratically I have such objects which get a String (we can call it "*strOrig*") in the constructor. As result it parse the String, and store all parsed informations in object's variables\attributes.
Obviously the String is checked to control if it fits some parameters before parsing starts. The static ArrayList contains further requisites (Strings which have to "match" with "*strOrig*" content to confirm it is valid) which are chosen by "the user" (I mean who implements these kind of classes).
It mean that I need to make the ArrayList modificable (both substituted by another ArrayList variable, or adding, changing and/or removing content for the existing one). I made it static, so it will not create a new ArrayList for every instantiate object.
So if List can't support add and similar, I think it's not my case. :-S
Maybe I could instantiate the ArrayList in an AbstractList object, so also Vector could also be used bye "the user"... but this is not very related.
So maybe private static AbstractList<String> strSearchs = new ArrayList<String>(Arrays.asList(" state changed from Full to ")); is my case...
Do you agree with me?
WalterLaan wrote:
ThePatcha wrote:
The variable is not intended to be Final, it can be modified via static methods, but I need it's static (so unique for every object instantiated) and it starts with a defaul value already inside.static means the opposite: it is the same for all instances, not unique per instance.My fault: I meant "it's always the same for all instanced objects". I used "unique" instead of "always the same", sorry.
Edited by: ThePatcha on Feb 2, 2010 6:45 AM