Different ways to copy an ArrayList

Question

Is there any difference in between these three, assuming a has type ArrayList<T>? (Aside from the fact that the compiler complains about unchecked operations in #3.)

1.

ArrayList<T> a1 = new ArrayList<T> (a);

2.

ArrayList<T> a2 = new ArrayList<T> ();
a2.addAll (a);

3.

ArrayList<T> a3 = (ArrayList<T>) (a.clone());

I'm willing to bet that #2 runs slower than #1 because of resizing issues, but I wouldn't know for sure. — Dennis Meng, Aug 15 '13 at 16:54
@DennisMeng Depends on if it allocates memory in a loop, or in one go. — S.D., Aug 15 '13 at 17:03
@S.D. True. I'd imagine it does allocate memory all at once (and resize once) as opposed to in a loop (where it resizes probably `log n` times or whatever) — Dennis Meng, Aug 15 '13 at 17:16

score 1 · Accepted Answer · edited May 23 '17 at 10:25

1

Cloning creates a new instance, holding the same elements. Clone works fine with Collections. So it is better not to use them.

ArrayList<T> a1 = new ArrayList<T> (a);

is a shallow copy and is comparatively faster.

This Thread may help you further

edited May 23 '17 at 10:25

Community

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answered Aug 15 '13 at 16:57

Rahul Tripathi

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Clone() works fine with Collections although it is considered broken. Correct me if I am wrong. :) – Rahul Tripathi Aug 15 '13 at 17:10
The javadoc for [ArrayList.clone](http://docs.oracle.com/javase/7/docs/api/java/util/ArrayList.html#clone()) says it makes a shallow copy; is the result any different from the constructor, #1? (Not counting performance issues.) – ajb Aug 15 '13 at 17:21
Ahh, misunderstood. Then "So it is better not to use them." relates to "creates a new instance". – Joop Eggen Aug 15 '13 at 18:51
I fully agree with you!! :) – Rahul Tripathi Aug 15 '13 at 18:51

Different ways to copy an ArrayList

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