Interview question

[S. Palanigounder]

When hashCode() is override, but not equals(), what will happen when object is used as the key in HashMap?

[Rob Spoor]

What do you think will happen? Will it find the bucket? If no, why not? If so, what will it do with the objects in that bucket?

[S. Palanigounder]

If hashCode is overwritten correctly, nothing will happen. The object can be the correct key used in HashMap.

[Michael Angstadt]

No, it needs both to work properly. As shown below, it calls the hashCode method when adding items to the Map, and calls both methods when retrieving items.

import java.util.*;
class Wrapper{
	public static void main(String args[]){
		Map<Wrapper, Integer> map = new HashMap<Wrapper, Integer>();
		map.put(new Wrapper("one"), 1);
		map.put(new Wrapper("two"), 2);
		System.out.println(map.get(new Wrapper("one")));
	}

	String string;
	Wrapper(String string){
		this.string = string;
	}
	
	public boolean equals(Object o){
		System.out.println(string + "--equals()");
		return string.equals(((Wrapper)o).string);
	}
	public int hashCode(){
		System.out.println(string + "--hashCode()");
		return string.hashCode();
	}
}

Prints the output:

one--hashCode()
two--hashCode()
one--hashCode()
one--equals()
1

[Henry Wong]

S. Palanigounder wrote:If hashCode is overwritten correctly, nothing will happen. The object can be the correct key used in HashMap.

Michael Angstadt wrote:No, it needs both to work properly.

I think this depends on what "overwritten correctly" and "work properly" means. If "overwritten correctly" just means that the hashCode / equals contract is not violated, then the object can be used as a key in a hashmap. And if "work properly" means that equality is based on instance, and not based on the value, then it should work properly. Of course, that could also have been achieved by not overriding hashCode() in the first place.

Henry

[S. Palanigounder]

Michael,

If you do not overwrite the equals method, which means using the default equals method, your code works fine.
The default equals method uses the bit-wise comparison.

[Michael Angstadt]

Henry Wong wrote:And if "work properly" means that equality is based on instance, and not based on the value, then it should work properly

Thanks Henry, yes, that's what I meant. I should have been more specific.

FYI: In my example, if you instead store the first instance of "new Wrapper("one")" in a variable and pass that to the Map.get() method, then you don't need the hashCode or the equals method.

S. Palanigounder wrote:The default equals method uses the bit-wise comparison.

No it doesn't...it compares the reference values for equality, as in:

return this == otherObj;

[S. Palanigounder]

Michael Angstadt wrote:

S. Palanigounder wrote:The default equals method uses the bit-wise comparison.

No it doesn't...it compares the reference values for equality, as in:

return this == otherObj;

What is the difference between "==" operator and "equals" method?

For example, String a=”12”; String b=”12”; String c= new String(“12”); They all have same content. But are they "==" or "equals"?

[Campbell Ritchie]

Have you tried it? What do you know about the behaviour of identical String literals, or copy constructors?

[Ulises Pulido]

Ok, many things could happen but having different scenarios.


Having hashCode well implemented means that the Object will find the correct bucket where this Object is in the Map buckets.

If you are using the same reference it will be found:

		  Map map = new HashMap();

		  ObjectWithHashCodeImplementedVeryWell object = new ObjectWithHashCodeImplementedVeryWell();
		  map.put(object, "Hello World");

		  String helloWorld = (String) map.get(object);

		  System.out.println(helloWorld);

this will print "Hello World"

If not it won't be found:

		  Map map = new HashMap();

		  ObjectWithHashCodeImplementedVeryWell object = new ObjectWithHashCodeImplementedVeryWell();
		  map.put(object, "Hello World");

		  ObjectWithHashCodeImplementedVeryWell object2 = new ObjectWithHashCodeImplementedVeryWell();
		  String helloWorld = (String) map.get(object2);

		  System.out.println(helloWorld);

this will print null


but if you had had equals overriden and well implemented it should return "Hello World"

[S. Palanigounder]

Ulises, Can you provide the override implementation for hashCode and equals that works?

[Campbell Ritchie]

Please look in books like Thinking in Java by Bruce Eckels, Effective Java by Joshua Bloch.
Please search for Angelika Langer's pages about the Java equals method.
Please search these fora because this question comes up frequently.

[Ulises Pulido]

S. Palanigounder wrote:Ulises, Can you provide the override implementation for hashCode and equals that works?

Well there is no equals and hascode implementation that can fit for all the objects.

Depending on the class you have created is the implementation you can do. What I can give you are the rules for the implementation.

If x == y then it must be that x.hashcode == y.hashcode
If x != y then x.hashcode might be == or != than y.hashcode
If x.equals(y) is true then it must be that x.hashcode == y.hashcode.
If !x.equals(y) then x.hashcode might be == or != than y.hashcode.
if x.equals(y) and y.equals(z) then x.equals(z). The same goes for hashcode.

Once you have implemented both methods and they behave as the rules stated before then you can say that you have a correct implementation of both methods.