First of all, for initializing a container you cannot use a primitive type (i.e. int; you can use int[] but as you want just an array of integers, I see no use in that). Instead, you should use Integer, as follows:

ArrayList<Integer> arl = new ArrayList<Integer>();

For adding elements, just use the add function:

arl.add(1);  
arl.add(22);
arl.add(-2);

Last, but not least, for printing the ArrayList you may use the build-in functionality of toString():

System.out.println("Arraylist contains: " + arl.toString());  

If you want to access the i element, where i is an index from 0 to the length of the array-1, you can do a :

int i = 0; // Index 0 is of the first element
System.out.println("The first element is: " + arl.get(i));

I suggest reading first on Java Containers, before starting to work with them.

Streams

1. In Java 8+ you can make a stream of your int array. Call either Arrays.stream or IntStream.of.
2. Call IntStream#boxed to use boxing conversion from int primitive to Integer objects.
3. Collect into a list using Stream.collect( Collectors.toList() ). Or more simply in Java 16+, call Stream#toList().

Example:

int[] ints = {1,2,3};
List<Integer> list = Arrays.stream(ints).boxed().collect(Collectors.toList());

In Java 16 and later:

List<Integer> list = Arrays.stream(ints).boxed().toList();

With streams added in Java 8 we can write code like:

int[] example1 = list.stream().mapToInt(i->i).toArray();
// OR
int[] example2 = list.stream().mapToInt(Integer::intValue).toArray();

Thought process:

• The simple Stream#toArray returns an Object[] array, so it is not what we want. Also, Stream#toArray(IntFunction<A[]> generator) which returns A[] doesn't do what we want, because the generic type A can't represent the primitive type int

• So it would be nice to have some kind of stream which would be designed to handle primitive type int instead of the reference type like Integer, because its toArray method will most likely also return an int[] array (returning something else like Object[] or even boxed Integer[] would be unnatural for int). And fortunately Java 8 has such a stream which is IntStream

• So now the only thing we need to figure out is how to convert our Stream<Integer> (which will be returned from list.stream()) to that shiny IntStream.

  Quick searching in documentation of Stream while looking for methods which return IntStream points us to our solution which is mapToInt(ToIntFunction<? super T> mapper) method. All we need to do is provide a mapping from Integer to int.

  Since ToIntFunction is functional interface we can also provide its instance via lambda or method reference.

  Anyway to convert Integer to int we can use Integer#intValue so inside mapToInt we can write:

  mapToInt( (Integer i) -> i.intValue() )
  

  (or some may prefer: mapToInt(Integer::intValue).)

  But similar code can be generated using unboxing, since the compiler knows that the result of this lambda must be of type int (the lambda used in mapToInt is an implementation of the ToIntFunction interface which expects as body a method of type: int applyAsInt(T value) which is expected to return an int).

  So we can simply write:

  mapToInt((Integer i)->i)
  

  Also, since the Integer type in (Integer i) can be inferred by the compiler because List<Integer>#stream() returns a Stream<Integer>, we can also skip it which leaves us with

  mapToInt(i -> i)