Java Array HashCode implementation

souLTower picture souLTower · Apr 13, 2009 · Viewed 50.3k times · Source

This is odd. A co-worker asked about the implementation of myArray.hashCode() in java. I thought I knew but then I ran a few tests. Check the code below. The odd thing I noticed is that when I wrote the first sys out the results were different. Note that it's almost like it's reporting a memory address and modifying the class moved the address or something. Just thought I would share.

int[] foo = new int[100000];
java.util.Random rand = new java.util.Random();

for(int a = 0; a < foo.length; a++) foo[a] = rand.nextInt();

int[] bar = new int[100000];
int[] baz = new int[100000];
int[] bax = new int[100000];
for(int a = 0; a < foo.length; a++) bar[a] = baz[a] = bax[a] = foo[a];

System.out.println(foo.hashCode() + " ----- " + bar.hashCode() + " ----- " + baz.hashCode() +  " ----- " + bax.hashCode());

// returns 4097744 ----- 328041 ----- 2083945 ----- 2438296
// Consistently unless you modify the class.  Very weird
// Before adding the comments below it returned this:
// 4177328 ----- 4097744 ----- 328041 ----- 2083945


System.out.println("Equal ?? " +
  (java.util.Arrays.equals(foo, bar) && java.util.Arrays.equals(bar, baz) &&
  java.util.Arrays.equals(baz, bax) && java.util.Arrays.equals(foo, bax)));

Answer

MahdeTo picture MahdeTo · Apr 13, 2009

The java.lang.Array hashCode method is inherited from Object, which means the hashcode depends on the reference. To get the hashcode based on the content of the array use Arrays.hashCode.

Beware though its a shallow hashcode implementation. A deep implementation is also present Arrays.deepHashCode.