Dear all,
Java's default contract (in Object) states that two objects with equal data
should return the same hashCode. In other words, if a.equals(b), their
return value of hashCode() must be the same.
Unfortunately, with Double values, new Double(0.0d).equals(new
Double(0.0d)) is false. This is because their internal representation
differs. Java's builtin Double thus returns a different hashCode for 0.0d
and 0.0d.
However, Apache's math library uses a mathematical comparison for
Complex/Dfp (perhaps also others), where 0.0d == 0.0d. This breaks the
contract, and thus causes problems when Complex or Dfp instances are used
in containers such as HashMap, HashSet, etc. See the bug report for more
details and test cases:
https://issues.apache.org/jira/browse/MATH1118
It is not quite clear how this should be fixed. Gilles posted a patch that
uses Java's equals to perform the comparison. This fixes the behavior
w.r.t. the contract but may surprise people who expect a mathematical
comparison.
Another possible fix (without a tentative patch at this time) would be to
change hashCode:
public int hashCode() {
if (imaginary == 0.0d && real == 0.0d) {
return ZERO.hashCode();
}
...
and similar for Dfp.
This fixes the issue for zero values. However, there may be issues with
normalized vs. unnormalized floating point values with the same
(mathematical) value but different internal representations, where this
kind of fix cannot be used. I'm not familiar enough with various floating
point implementations to know if Java always normalizes the values before
using them in a statement like hashCode().
Therefore, the suggestion above is mathematically nice for +/ 0.0d, but
otherwise much less safe than Gilles' patch.
Any comments? Probably best post them on JIRA:
https://issues.apache.org/jira/browse/MATH1118

Regards,
Cyrille Artho  http://artho.com/
Better once than never, for never too late.
 Shakespeare, "The Taming of the Shrew"

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