14 Sep
2003
14 Sep
'03
8:04 a.m.
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 \__/ qrczak@knm.org.pl ^^ http://qrnik.knm.org.pl/~qrczak/