Dnia nie 14. września 2003 01:04, Derek Elkins napisał:
> > > A...
> >
> > A (constructor), then ... (operator).
> > This is how I understand Haskell 98 lexing rules.
>
> My first thought was that it should produce, A.. ., as in (.) (A..), but
> obviously that would be wrong as A.. must be a function and therefore to
> be passed to (.) it would need to be (A..).
Argh, I was wrong. It's A.. (qualified operator), then . (operator).
So it's syntax error recognized during parsing.
> I take this to mean the (..) function from the Prelude module.
".." is a reserved operator, used for ranges.
--
__("< Marcin Kowalczyk
\__/ [EMAIL PROTECTED]
^^ http://qrnik.knm.org.pl/~qrczak/
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