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In Java, I'm trying to know whether an Integer is in an ArrayList<Integer>. I tried using this general solution, which warns that it doesn't work with arrays of primitives. That shouldn't be a problem, since Integers aren't primitives (ints are).

ArrayList<Integer> ary = new ArrayList<Integer>();
ary.add(2);

System.out.println(String.format("%s", Arrays.asList(ary).contains(2)));

Returns false. Any reason why? Any help is appreciated, although the less verbose the better.

Community
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JoseHdez_2
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    Why do you call `Arrays.asList(ary)`? `ary` was declared as an `ArrayList`? – bradimus Aug 19 '16 at 16:39
  • Sorry, I didn't realize `ArrayList` inherited from `List` and thus already had a `contains()` method... This was kind of a trivial question, so I'll leave it up to the community to decide whether to keep it or not. – JoseHdez_2 Aug 19 '16 at 16:47

3 Answers3

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Any reason why it returns false?

Simply because Arrays.asList(ary) will return a List<ArrayList<Integer>> and you try to find if it contains an Integer which cannot work.

As remainder here is the Javadoc of Arrays.asList(ary)

public static <T> List<T> asList(T... a) Returns a fixed-size list backed by the specified array.

Here as you provide as argument an ArrayList<Integer>, it will return a List of what you provided so a List<ArrayList<Integer>>.

You need to call List#contains(Object) on your list something like ary.contains(myInteger).

Nicolas Filotto
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You don't need asList() as ArrayList itself a list. Also you don't need String.format() to print the result. Just do in this way. This returns true : 

System.out.println(ary.contains(2));
Shahid
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The problem is that you're trying to coerce your ArrayList to a List when it already is. Arrays.asList takes an array or variable number of arguments and adds all of these to a list. All you need to do is call System.out.println(ary.contains(2));

Jacob
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