Objects as Map keys without Hashcode and equals.

[Ved Antani]

public class Contact
	{
		int i;
		String name;
		public Contact(int iVal, String nameVal)
		{
			i = iVal;
			name = nameVal;
		}
}	

public class MultiMap
{
	public static void main (String args[])
	{
		java.util.HashMap m = new java.util.HashMap();
		Contact m1 = new Contact(1, "name");
		Contact m2 = new Contact(1, "name");
		m.put(m1, "first");
		m.put(m2, "second");
		System.out.println(m.get(m1));
		System.out.println(m.get(m2));
	}

}

Output is:
first
second


How does this "get" method behave ? As both m1 and M2 have same values and I have not overridden hashcode(), will Object class's equals() method be called ?
I think the behaviour is like this:
1. There is no hashcode method so there is no way for the JVM to see if objects m1 and m2 contain different values
2. There is no equals method overridden so Object class's equals() is invoked and as both objects are different the code above works fine without m2 replacing m1's value.

[Campbell Ritchie]

I think your interpretation is correct

[Henry Wong]

Ved Antani wrote:

public class Contact
	{
		int i;
		String name;
		public Contact(int iVal, String nameVal)
		{
			i = iVal;
			name = nameVal;
		}
}	

public class MultiMap
{
	public static void main (String args[])
	{
		java.util.HashMap m = new java.util.HashMap();
		Contact m1 = new Contact(1, "name");
		Contact m2 = new Contact(1, "name");
		m.put(m1, "first");
		m.put(m2, "second");
		System.out.println(m.get(m1));
		System.out.println(m.get(m2));
	}

}

Output is:
first
second


How does this "get" method behave ? As both m1 and M2 have same values and I have not overridden hashcode(), will Object class's equals() method be called ?
I think the behaviour is like this:
1. There is no hashcode method so there is no way for the JVM to see if objects m1 and m2 contain different values
2. There is no equals method overridden so Object class's equals() is invoked and as both objects are different the code above works fine without m2 replacing m1's value.

1. There is no hashCode() method defined so it will be inherited from the Object class. This method will try to create a unique value per instance -- this value is also called the identity hash code.

Henry