[ruby-talk:01056] Re: ..

Hi,

In message "[ruby-talk:01054] .."
    on 00/01/05, Dave Thomas <[e-mail]> writes:

|  If I override '..', is it true that I can only use it in functional
|  form? The parser seems to have operator .. hard coded to be a Range.

Yes, and no.

  * operator .. and ... are hard coded, so you can only call it by
    using `send'.  So answer should be yes for the current
    interpreter.

  * in reference manual, .. and ... are listed among non-redfinable
    operators.  But the parser treats .. operator (not ... operator)
    as redfinable by accident.  It is bug.  It should not be able to
    override.  I'll fix it in next release.  So answer shoule be no.

The reasons that range operators are hard coded to be a Range are:

  * ../... operators appearing in conditionals, it is treated special,
    just like ../... operators in scalar context in Perl.  It cannot
    be done by methods.

  * since Range objects are immutable, range objects whose operands
    are literals will be cached in the parse tree.  It helps
    perfomance because ranges appear a lot for array slicing etc.

Clemens suggested that ../... operators out of conditionals to call
../... methods, but I'm hesitating it for the latter reason.

							matz.