I have created a default constructor which creates an empty "hand".
public Hand() {
hand = new ArrayList();
}
Whats the most efficient way to have a second constructor take an Array of cards, then add them a hand?
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2Most efficient will be to store the same reference if the given array - but there are a LOT of draw backs in it.amit– amit2012-11-14 18:51:41 +00:00Commented Nov 14, 2012 at 18:51
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In all games I know a hand does not contain many cards. Worrying about the most efficient way of doing this sound like premature optimization. I'd prefer the most clear/flexible way.madth3– madth32012-11-14 19:00:06 +00:00Commented Nov 14, 2012 at 19:00
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3 Answers
I would have one constructor to do both.
public Hand(Card... cards) {
hand = Arrays.asList(cards);
}
Or an ArrayList copy as Rohit Jain suggests.
6 Comments
Rohit Jain
Note:
Arrays.asList gives a fixed size ArrayList. of type: - java.util.Arrays.ArrayList
arne.b
But then he cannot add any cards to hand later, and the fact that he has a no-args constructor suggests that he would want to, doesn't it?
amit
@RohitJain: It should be fine - there is no reason why
hand won't be a List.Rohit Jain
@amit.. Actaully, problem is not with the type of
hand. Problem is that, it is unmodifiable list. That's why I used a constructor in my version.
Peter Lawrey
It's hard to know what the requirements are but this is the simplest. You might copy-on-write the
hand or want to make it mutable with a copy. |
You can do it like this: -
public Hand(String[] hands) {
hand = new ArrayList<String>(Arrays.asList(hands));
}
Or, you can iterate over your string array, and add individual elements to your ArrayList.
public Hand(String[] hands) {
hand = new ArrayList<String>();
for (String elem: hands)
hand.add(elem);
}
P.S: - Always declare a Generic Type List.
8 Comments
amit
This is definetly not most efficient
Rohit Jain
@amit.. Do you suggest iterating over the array and adding it to the
ArrayList?amit
No, I suggest a plain and stupid reference store or Arrays.asList alone (which fits perfectly assuming
hand is a List, as it should be)Rohit Jain
@amit.. Well, you know that
Arrays.asList returns a fixed size list right? So, if you just assign it, you won't be able to modify it.arne.b
@amit: If you think this is a good solution, why do you advice against it in your comment below the question?
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There's another option, Collections.addAll:
public Hand(Card[] cards) {
hand = new ArrayList<Card>();
Collections.addAll(hand, cards);
}
According to the documentation:
Adds all of the specified elements to the specified collection. Elements to be added may be specified individually or as an array. The behavior of this convenience method is identical to that of c.addAll(Arrays.asList(elements)), but this method is likely to run significantly faster under most implementations.
