Two methods for creating generic arrays

Question

I've learned following two methods for creating generic arrays.

One is

@SuppressWarnings("unchecked")
static <T> T[] array1(final Class<T> elementType, final int size) {

    return (T[]) Array.newInstance(elementType, size);
}

And the other is

static <T> T[] array2(final Class<T[]> arrayType, final int size) {

    return arrayType.cast(Array.newInstance(arrayType.getComponentType(), size));
}

Which is better? Are they same (internally)? Is any case actually wrong?

Following Question http://stackoverflow.com/questions/17226379/how-to-create-generic-primitive-arrays — Jin Kwon, Jun 21 '13 at 01:30

score 6 · Accepted Answer · answered Jun 20 '13 at 03:30

Behind the scenes, the two do effectively the same thing, except that in option 1 you're passing in T's class object and in option 2 you're passing in T[]'s class object.

I prefer option 1 because it's shorter and easier to read. Then again, it's the same as Array.newInstance with a cast added, so I'm not sure that your method adds a lot of value. :-)

newacct · Answer 2 · 2013-06-20T23:48:07.547

2

Note that the first one is not type-safe. For example, the following causes a ClassCastException:

array1(int.class, 5);

edited Jun 20 '13 at 23:48

answered Jun 20 '13 at 23:24

newacct

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Checked and confirmed. Both methods don't event compile for primitive types. – Jin Kwon Jun 21 '13 at 01:31
2

@JinKwon: No, the first one compiles and throws an exception. The second one doesn't compile, and is type-safe. – newacct Jun 21 '13 at 01:34
doesn't compile if assigned to an `int[]` variable – ZhongYu Jun 21 '13 at 01:55

score 0 · Answer 3 · answered Jun 21 '13 at 01:39

0

Another (better) way to create generic arrays

@SafeVarargs
static <E> E[] newArray(int length, E... array)
{
    return Arrays.copyOf(array, length);
}


Integer[] ints = newArray(10);

String[] strings = newArray(10);

answered Jun 21 '13 at 01:39

ZhongYu

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Note that this will fail if the caller itself is in generic scope and doesn't resolve `T` to a concrete type. – Paul Bellora Jun 21 '13 at 01:47
only because of erasure:) and the same problem exists for array1/array2() too. – ZhongYu Jun 21 '13 at 01:49
It doesn't because they use a `Class` type token and reflection to work around erasure. – Paul Bellora Jun 21 '13 at 02:06

Two methods for creating generic arrays

3 Answers3

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