There is a significant difference between: * A $ function without a type system * A statically checked $ function * A $ keyword without static checking
Curt Sampson wrote: > On 2009-09-30 13:45 -0300 (Wed), namekuseijin wrote: > > >> The Perl call is spot on. Specially because Haskell has been >> incorporating so much syntatic sugar that it's almost looking Perlish >> noise already: [examples deleted] >> > > No, I disagree with your particular examples; they're bog-standard > Haskell that don't use any syntatic sugar (. and $ are just library > functions), and I find them perfectly fine to read. Note that nothing > in there is inconsistent or interpreted in any sort of exceptional way, > unlike many things that look like that in Perl. > > It does take time to learn to read that sort of stuff, but once you've > got it, "simplifying" this sort of thing would only make it harder to > read, because it would be more verbose without saying anything more. > Haskell's concision is one of its most important strengths. > > (Incidently, a good exercise for learning to understand stuff like that > might be to go thorugh it and convert it to use parens instead of $, > full application instead of ., and so on.) > > cjs > -- Tony Morris http://tmorris.net/ _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
