SAX2 Namespace Support

David Megginson david at
Wed Jan 5 15:13:53 GMT 2000

John Aldridge <john.aldridge at> writes:

> I'm sorry, I don't see the problem.  The QNames are equal if ns() and
> name() both match.  The prefix is irrelevant.  This is what I suggested in
> my definition of the equality operator above.
>     if (qn1 == qn2) ...
> just does the right thing.  If some application really wants to compare the
> prefixed name, it can always write:
>     if (qn1.prefixedName () == qn2.prefixedName ()) ...
> I'm obviously missing something, since you and other knowledgable people
> don't regard this as self evidently obvious -- can you explain the problem
> to me, please?

In principle (the principle of least surprise), it's very bad
behaviour for two objects to be == in C++ or equals() in Java if any
of their publicly-accessible fields differ.  Think of sets, for

All the best,


David Megginson                 david at

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