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Harinder Bedi

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since Feb 16, 2013
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Recent posts by Harinder Bedi

Jesper de Jong wrote:There are some important things to note with the solutions given above:

Garrett's solution, with Arrays.asList() is efficient because it doesn't need to copy the content of the array. This method returns a List that is a "view" onto the array - a wrapper that makes the array look like a list. When you change an element in the list, the element in the original array is also changed. Note that the list is fixed size - if you try to add elements to the list, you'll get an exception.

Ernest's solution: new ArrayList(Arrays.asList(myArray)); copies the content of the array to a new ArrayList. The copy is ofcourse independent of the array, and you can add, remove etc. elements as you like.

Janarthan's solution, with Collections.addAll(myList, myStringArray); is essentially the same as Ernest's solution.

If you only need read access to the array as if it is a List and you don't want to add or remove elements from the list, then use Garrett's solution. Otherwise use Ernest's or Janarthan's solution.



Thanks for the info.
5 years ago