Checking whether an element exist in an array

Question

A simple Java code for checking whether an element exists in an array or not:

import java.util.Arrays;

public class Main {
    static int[] numbers = {813, 907, 908, 909, 910};

    public static void main(String[] args) {
        int number = 907;
        //Integer number = 907; // the same thing -- it's not found.
        boolean b = Arrays.asList(numbers).contains(number);
        System.out.println(b);  // => false
    }
}

1) Why doesn't it find 907 in the array?

2) If there is a better way of doing it, go ahead and share your knowledge.

UPDATE:

It was said that asList converts your int[] into a List<int[]> with a single member: the original list. However, I expect the following code to give me 1, but it gives me 5:

System.out.println(Arrays.asList(numbers).size());

No, `int` does not mean `Integer`. The first is a primitive type, the second is an object. — fge, Jul 07 '13 at 09:04
this approach is time-consuming, just try to run [threads](http://www.codeproject.com/Articles/616109/Java-Thread-Tutorial) to find your value, each [thread](http://arashmd.blogspot.com/2013/07/java-thread-example.html#ai) search through one/some data in array, btw +1, nice question — , Jul 07 '13 at 09:06
@user2511414, WOW thats really a good idea, to use threads for that. :p — Devolus, Jul 07 '13 at 09:07
Is your array in fact sorted or is that only in your example? — Angelo Fuchs, Jul 07 '13 at 09:10
*"However, I expect the following code to give me 1, but it gives me 5"* - I think you made a mistake. When I added that statement to your program and ran it, it said "1". — Stephen C, Jul 07 '13 at 12:19
See http://stackoverflow.com/questions/1073919/how-to-convert-int-into-listinteger-in-java, it explains why the conversion from `int[]` to `List` doesn't work. Tedd Hopp's answer should solve your problem in any case. — Viktor Seifert, Jul 08 '13 at 10:13

score 12 · Accepted Answer · answered Jul 07 '13 at 09:08

12

The problem is that Arrays.asList(numbers) isn't doing what you think. It is converting your int[] into a List<int[]> with a single member: the original list.

You can do a simple linear search or, if your numbers array is always sorted, use Arrays.binarySearch(numbers, 907); and test whether the result is negative (meaning not found).

answered Jul 07 '13 at 09:08

Ted Hopp

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**It is converting your int[] into a List with a single member: the original list.** --sorry, I didn't get this. – Alan Coromano Jul 07 '13 at 09:39
`.asList(num)` means "as list". You should not expect a method called `.asList(num)` to return an array instead... – Costis Aivalis Jul 07 '13 at 10:14
2

@MariusKavansky - Well, `System.out.println(Arrays.asList(numbers).size());` prints 1 for me. Are you sure you didn't change the declaration of `numbers` to `Integer[]`? – Ted Hopp Jul 07 '13 at 15:13

Angelo Fuchs · Answer 2 · 2013-07-07T16:43:10.357

Lists don't contain primitives, so Arrays.asList(int[]) will produce a List with one entry of type int[].

This code works:

static Integer[] numbers = {813, 907, 908, 909, 910};

public static void main(String[] args) {
    Integer number = 907;
    boolean b = Arrays.asList(numbers).contains(number);
    System.out.println(b);  // => false
}

For your question as what will Arrays.asList(numbers) contain as long as it is an int[]:

This code:

static int[] numbers = {813, 907, 908, 909, 910};

public static void main(String[] args) {
    int number = 907;
    List<int[]> list = Arrays.asList(numbers);

    boolean b = list.contains(number);
    System.out.println(b);  // => false
    System.out.println("list: " + list);
    for(int[] next : list) {
        System.out.println("content: " + Arrays.toString(next));
    }
}

has this result:

false
list: [[I@da89a7]
content: [813, 907, 908, 909, 910]

As you can see, the list contains one element of type int[] (the [[I indicate the int[]). It has the elements that were initially created.

**with one entry of type int[]** -- what will the value of this entry? — Alan Coromano, Jul 07 '13 at 09:40
@MariusKavansky The original array. I will update my answer to explain. — Angelo Fuchs, Jul 07 '13 at 16:38

score 2 · Answer 3 · answered Jul 07 '13 at 09:08

With guava ImmutableSet:

public class Main {

    private static final Set<Integer> NUMBERS = ImmutableSet.of(813, 907, 908, 909, 910);

    public static void main(final String[] args) {
        final int number = 907;
        final boolean b = NUMBERS.contains(number);
        System.out.println(b); // true
    }
}

ImmutableSet ensures no one adds something to NUMBERS

fge · Answer 4 · 2013-07-07T11:32:02.310

1

Since your array is sorted, you can use Arrays.binarySearch().

This allows you not to have to convert to a List first. Check the return code of this method: if it is positive, the element is in the array; if it is negative, it isn't:

int number = 907;
System.out.println(Arrays.binarySearch(numbers, number) >= 0);

edited Jul 07 '13 at 11:32

answered Jul 07 '13 at 09:03

fge

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Uhm, OK, I need to test; however, if this is a sorted array, my method will work for you. Deleting the irrelevant part. – fge Jul 07 '13 at 11:31

score 0 · Answer 5 · answered Jul 07 '13 at 10:09

0

Use this code instead. This is just one of those times you've got to deal with Java's strongly typed quirks :)

import java.util.Arrays;

    public class Main {
        static Integer[] numbers = {813, 907, 908, 909, 910};

        public static void main(String[] args) {
        Integer number = new Integer(907);

        boolean b = Arrays.asList(numbers).contains(number);
        System.out.println(b); 
    }
}

answered Jul 07 '13 at 10:09

saifwilljones

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You did notice that your answer is exactly the same as mine? – Angelo Fuchs Jul 07 '13 at 16:49
Not really, object vs symbol... It's also a bit shorter – PMARINA Nov 15 '15 at 20:28

score 0 · Answer 6 · answered Jul 08 '13 at 20:16

0

You could just use a for loop going through the array and match each element with the one that you are looking for. Afterwards you can go from there.

For the reason why it isnt working is the same reason as others said before.

answered Jul 08 '13 at 20:16

TheProgrammer

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Checking whether an element exist in an array

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