What's the reason I can't create generic array types in Java?

Question

What's the reason why Java doesn't allow us to do

private T[] elements = new T[initialCapacity];

I could understand .NET didn't allow us to do that, as in .NET you have value types that at run-time can have different sizes, but in Java all kinds of T will be object references, thus having the same size (correct me if I'm wrong).

What is the reason?

Answer 1

The main reason is due to the fact that arrays in Java are covariant.

There's a good overview here.

Answer 2

By failing to provide a decent solution, you just end up with something worse IMHO.

The common work around is as follows.

T[] ts = new T[n];

is replaced with (assuming T extends Object and not another class)

T[] ts = (T[]) new Object[n];

I prefer the first example, however more academic types seem to prefer the second, or just prefer not to think about it.

Most of the examples of why you can't just use an Object[] equally apply to List or Collection (which are supported), so I see them as very poor arguments.

Note: this is one of the reasons the Collections library itself doesn't compile without warnings. If this use-case cannot be supported without warnings, something is fundamentally broken with the generics model IMHO.

Answer 3

class can declare an array of type T[], but it cannot directly instantiate such an array. Instead, a common approach is to instantiate an array of type Object[], and then make a narrowing cast to type T[], as shown in the following:

  public class Portfolio<T> {
  T[] data;
 public Portfolio(int capacity) {
   data = new T[capacity];                 // illegal; compiler error
   data = (T[]) new Object[capacity];      // legal, but compiler warning
 }
 public T get(int index) { return data[index]; }
 public void set(int index, T element) { data[index] = element; }
}

Answer 4

T vals[]; // OK

But, you cannot instantiate an array of T // vals = new T[10]; // can't create an array of T

The reason you can’t create an array of T is that there is no way for the compiler to know what type of array to actually create.

Answer 5

Try this:

List<?>[] arrayOfLists = new List<?>[4];

Answer 6

As others already mentioned, you can of course create via some tricks.

But it's not recommended.

Because the type erasure and more importantly the covariance in array which just allows a subtype array can be assigned to a supertype array, which forces you to use explicit type cast when trying to get the value back causing run-time ClassCastException which is one of the main objectives that generics try to eliminate: Stronger type checks at compile time.

Object[] stringArray = { "hi", "me" };
stringArray[1] = 1;
String aString = (String) stringArray[1]; // boom! the TypeCastException

A more direct example can found in Effective Java: Item 25.

---

covariance: an array of type S[] is a subtype of T[] if S is a subtype of T

Answer 7

From Oracle tutorial:

You cannot create arrays of parameterized types. For example, the following code does not compile:

List<Integer>[] arrayOfLists = new List<Integer>[2];  // compile-time error

The following code illustrates what happens when different types are inserted into an array:

Object[] strings = new String[2];
strings[0] = "hi";   // OK
strings[1] = 100;    // An ArrayStoreException is thrown.

If you try the same thing with a generic list, there would be a problem:

Object[] stringLists = new List<String>[];  // compiler error, but pretend it's allowed
stringLists[0] = new ArrayList<String>();   // OK
stringLists[1] = new ArrayList<Integer>();  // An ArrayStoreException should be thrown,
                                            // but the runtime can't detect it.

If arrays of parameterized lists were allowed, the previous code would fail to throw the desired ArrayStoreException.

To me, it sounds very weak. I think that anybody with a sufficient understanding of generics, would be perfectly fine, and even expect, that the ArrayStoredException is not thrown in such case.

Answer 8

If we cannot instantiate generic arrays, why does the language have generic array types? What's the point of having a type without objects?

The only reason I can think of, is varargs - foo(T...). Otherwise they could have completely scrubbed generic array types. (Well, they didn't really have to use array for varargs, since varargs didn't exist before 1.5. That's probably another mistake.)

So it is a lie, you can instantiate generic arrays, through varargs!

Of course, the problems with generic arrays are still real, e.g.

static <T> T[] foo(T... args){
    return args;
}
static <T> T[] foo2(T a1, T a2){
    return foo(a1, a2);
}

public static void main(String[] args){
    String[] x2 = foo2("a", "b"); // heap pollution!
}

We can use this example to actually demonstrate the danger of generic array.

On the other hand, we've been using generic varargs for a decade, and the sky is not falling yet. So we can argue that the problems are being exaggerated; it is not a big deal. If explicit generic array creation is allowed, we'll have bugs here and there; but we've been used to the problems of erasure, and we can live with it.

And we can point to foo2 to refute the claim that the spec keeps us from the problems that they claim to keep us from. If Sun had more time and resources for 1.5, I believe they could have reached a more satisfying resolution.

Answer 9

In my case, I simply wanted an array of stacks, something like this:

Stack<SomeType>[] stacks = new Stack<SomeType>[2];

Since this was not possible, I used the following as a workaround:

1. Created a non-generic wrapper class around Stack (say MyStack)
2. MyStack[] stacks = new MyStack[2] worked perfectly well

Ugly, but Java is happy.

Note: as mentioned by BrainSlugs83 in the comment to the question, it is totally possible to have arrays of generics in .NET

Answer 10

The reason this is impossible is that Java implements its Generics purely on the compiler level, and there is only one class file generated for each class. This is called Type Erasure.

At runtime, the compiled class needs to handle all of its uses with the same bytecode. So, new T[capacity] would have absolutely no idea what type needs to be instantiated.

Answer 11

It is because generics were added on to java after they made it, so its kinda clunky because the original makers of java thought that when making an array the type would be specified in the making of it. So that does not work with generics so you have to do E[] array=(E[]) new Object[15]; This compiles but it gives a warning.

Answer 12

I like the answer indirectly given by Gafter. However, I propose it is wrong. I changed Gafter's code a little. It compiles and it runs for a while then it bombs where Gafter predicted it would

class Box<T> {

    final T x;

    Box(T x) {
        this.x = x;
    }
}

class Loophole {

    public static <T> T[] array(final T... values) {
        return (values);
    }

    public static void main(String[] args) {

        Box<String> a = new Box("Hello");
        Box<String> b = new Box("World");
        Box<String> c = new Box("!!!!!!!!!!!");
        Box<String>[] bsa = array(a, b, c);
        System.out.println("I created an array of generics.");

        Object[] oa = bsa;
        oa[0] = new Box<Integer>(3);
        System.out.println("error not caught by array store check");

        try {
            String s = bsa[0].x;
        } catch (ClassCastException cause) {
            System.out.println("BOOM!");
            cause.printStackTrace();
        }
    }
}

The output is

I created an array of generics.
error not caught by array store check
BOOM!
java.lang.ClassCastException: java.lang.Integer cannot be cast to java.lang.String
    at Loophole.main(Box.java:26)

So it appears to me you can create generic array types in java. Did I misunderstand the question?

Answer 13

The answer was already given but if you already have an Instance of T then you can do this:

T t; //Assuming you already have this object instantiated or given by parameter.
int length;
T[] ts = (T[]) Array.newInstance(t.getClass(), length);

Hope, I could Help, Ferdi265

Answer 14

There surely must be a good way around it (maybe using reflection), because it seems to me that that's exactly what ArrayList.toArray(T[] a) does. I quote:

public <T> T[] toArray(T[] a)

Returns an array containing all of the elements in this list in the correct order; the runtime type of the returned array is that of the specified array. If the list fits in the specified array, it is returned therein. Otherwise, a new array is allocated with the runtime type of the specified array and the size of this list.

So one way around it would be to use this function i.e. create an ArrayList of the objects you want in the array, then use toArray(T[] a) to create the actual array. It wouldn't be speedy, but you didn't mention your requirements.

So does anyone know how toArray(T[] a) is implemented?

Answer 15

It's because Java's arrays (unlike generics) contain, at runtime, information about its component type. So you must know the component type when you create the array. Since you don't know what T is at runtime, you can't create the array.

Answer 16

Quote:

Arrays of generic types are not allowed because they're not sound. The problem is due to the interaction of Java arrays, which are not statically sound but are dynamically checked, with generics, which are statically sound and not dynamically checked. Here is how you could exploit the loophole:

class Box<T> {
    final T x;
    Box(T x) {
        this.x = x;
    }
}

class Loophole {
    public static void main(String[] args) {
        Box<String>[] bsa = new Box<String>[3];
        Object[] oa = bsa;
        oa[0] = new Box<Integer>(3); // error not caught by array store check
        String s = bsa[0].x; // BOOM!
    }
}

We had proposed to resolve this problem using statically safe arrays (aka Variance) bute that was rejected for Tiger.

-- gafter

(I believe it is Neal Gafter, but am not sure)

See it in context here: http://forums.sun.com/thread.jspa?threadID=457033&forumID=316