How to find the index of an element in an array in Java?

Question

In this case, you could create e new String from your array of chars and then do an indeoxOf("e") on that String:

System.out.println(new String(list).indexOf("e")); 

But in other cases of primitive data types, you'll have to iterate over it.

Alternatively, you can use Commons Lang ArrayUtils class:

int[] arr = new int{3, 5, 1, 4, 2};
int indexOfTwo = ArrayUtils.indexOf(arr, 2);

There are overloaded variants of indexOf() method for different array types.

That's not even valid syntax. And you're trying to compare to a string. For arrays you would have to walk the array yourself:

public class T {
  public static void main( String args[] ) {
    char[] list = {'m', 'e', 'y'};

    int index = -1;

    for (int i = 0; (i < list.length) && (index == -1); i++) {
        if (list[i] == 'e') {
            index = i;
        }
    }

    System.out.println(index);
  }
}

If you are using a collection, such as ArrayList<Character> you can also use the indexOf() method:

ArrayList<Character> list = new ArrayList<Character>();
list.add('m');
list.add('e');
list.add('y');

System.out.println(list.indexOf('e'));

There is also the Arrays class which shortens above code:

List list = Arrays.asList(new Character[] { 'm', 'e', 'y' });
System.out.println(list.indexOf('e'));

For primitive arrays

Starting with Java 8, the general purpose solution for a primitive array arr, and a value to search val, is:

public static int indexOf(char[] arr, char val) {
    return IntStream.range(0, arr.length).filter(i -> arr[i] == val).findFirst().orElse(-1);
}

This code creates a stream over the indexes of the array with IntStream.range, filters the indexes to keep only those where the array's element at that index is equal to the value searched and finally keeps the first one matched with findFirst. findFirst returns an OptionalInt, as it is possible that no matching indexes were found. So we invoke orElse(-1) to either return the value found or -1 if none were found.

Overloads can be added for int[], long[], etc. The body of the method will remain the same.

For Object arrays

For object arrays, like String[], we could use the same idea and have the filtering step using the equals method, or Objects.equals to consider two null elements equal, instead of ==.

But we can do it in a simpler manner with:

public static <T> int indexOf(T[] arr, T val) {
    return Arrays.asList(arr).indexOf(val);
}

This creates a list wrapper for the input array using Arrays.asList and searches the index of the element with indexOf.

This solution does not work for primitive arrays, as shown here: a primitive array like int[] is not an Object[] but an Object; as such, invoking asList on it creates a list of a single element, which is the given array, not a list of the elements of the array.

I believe the only sanest way to do this is to manually iterate through the array.

for (int i = 0; i < list.length; i++) {
  if (list[i] == 'e') {
    System.out.println(i);
    break;
  }
}