Re: Did the sort do anything?
On 5/16/2011 5:02 AM, Lawrence D'Oliveiro wrote:
In message<kemdnZVnBva8n0zQnZ2dnUVZ_uqdnZ2d@earthlink.com>, Patricia
Shanahan wrote:
Distinct objects probably have different identity hash codes regardless of
whether their fields all have the same values.
???By contract, any two objects for which equals(Object) returns true must
return the same hash code value.???
System.identityHashCode(x) does not necessarily return the same as
x.hashCode(), and usually will not unless x inherits equals and hashCode
from Object. The contracts for System.identityHashCode(x) and
x.hashCode() match if, and only if, x.equals(y) is equivalent to x==y.
<http://developer.android.com/reference/java/lang/Object.html#hashCode%28%29>
???Distinct??? would presumably mean ???not equal???.
It means "not the same object".
For example, equality for Integer is based only on the numeric value of
the integer the object represents. Integer objects are Comparable, with
sort order the same as numeric order of the represented integer.
Integer should be the poster child for not caring about sort stability.
This program:
public class IdentityHashCode {
public static void main(String[] args) {
Integer x = new Integer(3);
Integer y = new Integer(3);
System.out.printf("x.equals(y)=%b, x.compareTo(y)=%d, "
+ "x.hashCode()=%d "
+ "System.identityHashCode(x)=%d%n", x.equals(y), x
.compareTo(y), x.hashCode(), System.identityHashCode(x));
System.out.printf("y.equals(x)=%b, y.compareTo(x)=%d, "
+ "y.hashCode()=%d "
+ "System.identityHashCode(y)=%d%n", y.equals(x), y
.compareTo(x), y.hashCode(), System.identityHashCode(y));
}
}
produces as typical output:
x.equals(y)=true, x.compareTo(y)=0, x.hashCode()=3
System.identityHashCode(x)=4072869
y.equals(x)=true, y.compareTo(x)=0, y.hashCode()=3
System.identityHashCode(y)=11077203
Whether the sort algorithm is stable or not can affect subsequent
program behavior even for sorting Integer objects by their natural order.
Patricia