Initialize ArrayList<Long>

Question

Why I can initialize ArrayList, like this:

ArrayList<Integer> x = new ArrayList<Integer>(Arrays.asList(1,2));

But got Error when using:

ArrayList<Long> x = new ArrayList<Long>(Arrays.asList(1,2));

Answer 1

You have to specify a Long number using literal l or L.

List<Long> x = new ArrayList<Long>(Arrays.asList(1L, 2L));

Otherwise 1 and 2 are understood as integers resulting in List<Integer>.

Answer 2

That's because 1 and 2 are ints and Arrays.asList(1, 2) creates a List<Integer>.

And the copy constructor of ArrayList requires the argument to be of the same generic type.

You have several options but the simplest one is to change the ints into longs by adding a L suffix:

List<Long> x = new ArrayList<Long>(Arrays.asList(1L, 2L));

Note that with Java 9 you can also write:

List<Long> x = List.of(1L, 2L);

Answer 3

Explanation

Java automatically transforms int to long if needed.

However, Java does not do the same if a transformation from Integer to Long is needed.

The function Arrays.asList(...) returns a List<E> with E being the type used as parameters. As you use 1, 2, 3 the type is int. However the generic usage of data-types as List<int> is not possible in Java (at least currently). Therefore it automatically transforms int to Integer and produces a List<Integer> object. This process is called auto-boxing, Java can do this for all data-types to their corresponding object representation.

If you now use the constructor new ArrayList<Integer>(List<E> list) it expects E to be something of type Integer. So a List<Integer> works as input.

But when you use new ArrayList<Long>(List<E> list) obviously E needs to be of type Long. However the object Integer is not of type Long thus it does not accept the parameter. The first common type of Integer and Long is the abstract class Number (which also holds Double, Float and others) (documentation).

---

Solution

So it all revolves around the input 1, 2, 3 being interpreted as int instead of long. You can fix this by explicitly telling Java to interpret the numbers as long, you do so by appending l or L after the number:

new ArrayList<Long>(Arrays.asList(1L, 2L, 3L));

Now you receive a List<Long> which then is added to an ArrayList<Long>.

---

Note that the same technique can be used to explicitly interpret decimal numbers as float instead of double: 1.5F or 1.5f

Answer 4

Because Arrays.asList(1,2) will implicitly return a List<Integer>.

You can fix this by using the following idiom:

ArrayList<Long> x = new ArrayList<Long>(Arrays.asList(1l,2l));