On Mon, Jul 18, 2011 at 11:52:15PM -0700, Ting Lei wrote: > > Thanks, Brandon, > > That's a very clear explanation. I remembered that it's low, but forgot that > it's this low. > Maybe it's because of my mis-understanding that in C, infix operators have > lower precedence. > > Just curious, the following is not allowed in Haskell either for the same > reason. > > applySkip i f ls = (take i) ls ++ $ f $ drop i ls > > I read somewhere that people a couple of hundreds of years ago can manage to > express things using ($)-like notation without any parenthesis at all. > Is it possible to define an operator to achieve this which works > with infix operators? If so, then ($) and parentheses are then > really equivalent.
($) and parentheses are quite different. Anyone who tells you they are equivalent, although probably well-meaning, should be ignored. Here is everything you need to know about ($): infixr 0 $ f $ x = f x Although working out all the implications of this definition may take a bit of practice. -Brent _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
