Creating generic array in Java

Question

I got "Cannot create a generic array of T" while trying to create a generic array like this

tmp = new T[n];

I found [this][1] [1]: https://stackoverflow.com/questions/529085/how-to-create-a-generic-array-in-java, but having to pass class every time I want to call method seemed too much trouble to me.

So I solved it in the next way

this.tmp=Arrays.copyOf(list, n);

Is this a good solution considering the rest of code? Why yes? Why no?

Full code:

public class MergeSort<T extends Comparable<? super T>>{
    private T[] A;
    private T[] tmp;
    private int n;
    
    public void sort(T[] list){
        this.A=list;
        this.n=list.length;
        this.tmp=Arrays.copyOf(list, n);
        mergeSort(0,n-1);
    }
    
    private void mergeSort(int low,int high){
        if(low<high){
            int middle=(low+high)/2;
            mergeSort(low, middle);
            mergeSort(middle+1, high);
            merge(low,middle,high);
        }
    }
    private void merge(int left,int right,int rightEnd){
        int leftEnd=right,k=left;
        right++;
        while(left<=leftEnd && right<=rightEnd){
            if(A[left].compareTo(A[right])<=0)
                tmp[k++]=A[left++];
            else
                tmp[k++]=A[right++];
        }
        
        while(left<=leftEnd)
            tmp[k++]=A[left++];
        while(right<=rightEnd)
            tmp[k++]=A[right++];
        
        for (int i = 0; i <= rightEnd; i++) {
              A[i] = tmp[i];
        }
    }
}

Maybe a better question would be; what if I used following line of code instead?

this.tmp=(T[]) Arrays.copyOf(new Object[]{}, n)

and by the way, your merge sort is broken http://googleresearch.blogspot.co.uk/2006/06/extra-extra-read-all-about-it-nearly.html — Sleiman Jneidi, Dec 27 '15 at 22:35
@Sleiman Thank for pointing out bug. As for copying, yes here i need to copy array anyway, but this is where i started thinking about how to create generic array. Maybe better question would be if way around like this 'this.tmp=(T[]) Arrays.copyOf(new Object[]{}, n);' would be very bad solution and why. — redMist, Dec 27 '15 at 23:00
In Java, every array has a public `clone()` method. You could simply write `this.tmp = list.clone();`. — VGR, Jan 02 '16 at 03:25

score 2 · Accepted Answer · answered Dec 27 '15 at 22:41

2

You can use this method from guava

  static <T> T[] newArray(T[] reference, int length) {
    Class<?> type = reference.getClass().getComponentType();
    @SuppressWarnings("unchecked")
    T[] result = (T[]) Array.newInstance(type, length);
    return result;
  }

answered Dec 27 '15 at 22:41

v.ladynev

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I'll use this method as it seems a preferable solution, but I can't mark it as answer, thanks though. – redMist Dec 27 '15 at 22:59

Creating generic array in Java

1 Answers1