How to initialize generic array?

Question

This method is used to split array into chunks reference. I want to make this method generic.

Problem is, I can not initialize array like this.

T[][] arrays = new T[chunks][];

complete method

 public <T> T[][] splitArray(T[] arrayToSplit, int chunkSize) {
        if (chunkSize <= 0) {
            return null;
        }
        int rest = arrayToSplit.length % chunkSize;
        int chunks = arrayToSplit.length / chunkSize + (rest > 0 ? 1 : 0);
        T[][] arrays = new T[chunks][];
        for (int i = 0; i < (rest > 0 ? chunks - 1 : chunks); i++) {
            arrays[i] = Arrays.copyOfRange(arrayToSplit, i * chunkSize, i * chunkSize + chunkSize);
        }
        if (rest > 0) {
            arrays[chunks - 1] = Arrays.copyOfRange(arrayToSplit, (chunks - 1) * chunkSize, (chunks - 1) * chunkSize + rest);
        }
        return arrays;
    }

What is correct way to initialize generic array?

score 7 · Accepted Answer · answered May 10 '18 at 05:55

7

You cannot initialise arrays with a generic parameter. That is a restriction on generics.

A workaround is to create an Object[][] and cast it to T[][]:

T[][] arrays = (T[][])new Object[chunks][];

answered May 10 '18 at 05:55

Sweeper

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Wrong 1st argument type. Found: 'byte[]', required: 'T[]' – Khemraj Sharma May 10 '18 at 05:59
Why does this show error when invoking this method. – Khemraj Sharma May 10 '18 at 05:59
Can you please tell syntax to call this method also. – Khemraj Sharma May 10 '18 at 06:01
3

@Khemraj because your method is generic, you can't call it with primitive types: https://docs.oracle.com/javase/tutorial/java/generics/restrictions.html#instantiate Use a `Byte[]` (note the capital letter). – Sweeper May 10 '18 at 06:04
Thanks for your help with useful links, i will sure study generic docs to improve my understandings. – Khemraj Sharma May 10 '18 at 06:06
Thanks @Sweeper because of this answer I check in ArrayList implementation it also more like this. :) – janith1024 May 10 '18 at 06:08

score 3 · Answer 2 · answered May 10 '18 at 08:37

3

You can't create a generic array, but you can declare a generic array.

T[] test = null; // works
T[] test2 = new T[10]; // fails
T[] test3 = (T[]) new Object[10]; // works

At the same time, you should be careful with this

answered May 10 '18 at 08:37

Eugene

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is there any way to declare a generic array without the java compiler logging "uses unchecked or unsafe operations." when compiling the file that does this? `test3` causes that warning – notacorn Aug 20 '20 at 03:41
1

@notacorn no, since you can not declare a generic array, you will _always_ have the cast to `T[]`, always. – Eugene Aug 20 '20 at 10:43

score 1 · Answer 3 · answered May 10 '18 at 06:10

You can't make a generic array directly since type variable information is lost at runtime due to erasure.

However, in your case there is a workaround, since the output type of the method is just an array of type arrayToSplit and arrays in Java do have their type information available at runtime.

So instead of:

T[][] arrays = new T[chunks][];

you can do:

T[][] arrays = (T[][])Array.newInstance(arrayToSplit.getClass(), chunks);

Mixing arrays and generics can be confusing and error-prone though. If possible, I would use the collections API where possible and use List<T> instead of arrays.

Guava even has a method that I think does exactly what you want:

Lists.partition(List, int)

Returns consecutive sublists of a list, each of the same size (the final list may be smaller). For example, partitioning a list containing [a, b, c, d, e] with a partition size of 3 yields [[a, b, c], [d, e]] -- an outer list containing two inner lists of three and two elements, all in the original order.

How to initialize generic array?

3 Answers3