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hashcode() returns memory address ?

 
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import java.io.*;

public class test1
{
public static void main(String[] args)
{
String x = "raja";
String y = "raja";

System.out.println("hello "+x.hashCode()+" "+y.hashCode() );

Integer i1 = new Integer(10);
Integer i3 = i1;
System.out.println("hello1 "+i1.hashCode()+" "+i3.hashCode() );

Float f1 = new Float(10.1);
Float f2 = new Float(10.1);
System.out.println("hello1 "+f1.hashCode()+" "+f2.hashCode() );
}
----
above example gave output as
hello 3492774 3492774
hello1 10 10
hello1 1092721050 1092721050
Here why hashcode value of Integers is same as the Integer Value?
 
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Here are the rules for the hashcode function (as laid out in the API Spec for Object):


The general contract of hashCode is:
- Whenever it is invoked on the same object more than once during an execution of a Java application, the hashCode method must consistently return the same integer, provided no information used in equals comparisons on the object is modified. This integer need not remain consistent from one execution of an application to another execution of the same application.
- If two objects are equal according to the equals(Object) method, then calling the hashCode method on each of the two objects must produce the same integer result.
- It is not required that if two objects are unequal according to the equals(java.lang.Object) method, then calling the hashCode method on each of the two objects must produce distinct integer results. However, the programmer should be aware that producing distinct integer results for unequal objects may improve the performance of hashtables.


So, the hashcode must always return the same value if two objects are equal (per their equals method) and it must return an int. It seems to me that the int value wrapped by this Integer is the perfect choice to use as a hashcode.
Corey
 
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