Java collections - overriding equals and hashCode

Question

class Hash {
  int a;

  Hash(int h){
    a=h;
  }

  public boolean equals(Object o) {     
    Boolean h=super.equals(o);
    System.out.println("Inside equals ");
    return h;
  }

  public int hashCode() {    
    System.out.println("Inside Hash");    
    return 2;
  }    
}

public class Eq {    
  public static void main(String...r) {    
    HashMap<Hash,Integer> map=new HashMap<Hash,Integer>();    
    Hash j=new Hash(2);    
    map.put(j,1);
    map.put(j,2);
    System.out.println(map.size());
  }
}

output was

inside hash

inside hash
1

Since it returns the same hashcode , the second time an object is added in hashmap it must use the equals method but it doesnt call . So wats the problem here?

stackoverflow.com/editing-help

Matt Ball
– Matt Ball

2011-07-10 13:41:11 +00:00
Commented Jul 10, 2011 at 13:41 — Matt Ball
– Matt Ball, Commented Jul 10, 2011 at 13:41
Overriding equals and hashCode in Java

Matt Ball
– Matt Ball

2011-07-10 13:44:45 +00:00
Commented Jul 10, 2011 at 13:44 — Matt Ball
– Matt Ball, Commented Jul 10, 2011 at 13:44

highlycaffeinated · Accepted Answer · 2011-07-10 14:27:34Z

7

The HashMap is testing with == before .equals, and since you are putting the same object twice, the first test passes. Try with:

    Hash j=new Hash(2);
    Hash k=new Hash(2);
    map.put(j,1);
    map.put(k,2);

edited Jul 10, 2011 at 14:27

answered Jul 10, 2011 at 13:46

highlycaffeinated

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1 Comment

M Platvoet Over a year ago

Relevant code at line 393

Howard · Accepted Answer · 2011-07-10 13:47:59Z

0

The equality check is done by HashMap in three steps:

hash code is different => unequal
objects are identical (==) => equal
equal method gives true => equal

The second step prevents calling equals since identical objects are always assumed equal.

answered Jul 10, 2011 at 13:47

Howard

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Comments

BlueDog · Accepted Answer · 2011-07-10 13:45:09Z

-3

From the doc

put(): Associates the specified value with the specified key in this map. If the map previously contained a mapping for this key, the old value is replaced.

answered Jul 10, 2011 at 13:45

BlueDog

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3 Comments

BlueDog Over a year ago

The "specified key" is the hashcode, isn't it? Since the key is already used at the time of the second put, there is no need to check for equality hence no equals() call. Or did I mess it up?

Petar Minchev Over a year ago

The correct answer is the first one - there is a link to the hashmap code.

g051051 Over a year ago

HashMap supports key collisions by creating an internal list containing all items with a specific hashcode. On a put, it searches all items with the same hashcode and uses == || .equals to make the final determination. so in the original question, it's hitting the hashCode, but the == comaprison is succeeding and short-circuitng the call to .equals.

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