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