Hi Mark, > Suppose I have functions > > f :: Int -> Int > f x -> x * x
I suppose you mean: f x = x * x > g :: Int -> Int > g x -> x + 1 > > The lazy application operator "$" allows me to do: > > f $ g x > > instead of > > f (g x) > > But I don't understand why it works this way! Let me explain. > f is a function, and application has highest precedence, so unless > it sees a bracket, it should take the next thing it sees as an > argument. Yes, but "$" cannot be an argument. In the Haskell grammar for expressions ( http://www.haskell.org/onlinereport/exps.html ) an application (fexp) consists of one or aexp's and an aexp cannot be an operator (at least not, without parentheses around it). A simpler way to see this is to write application as an explicit operator. Let's call it @. Above expression then reads f $ g @ x And @ binds stronger than $, alas f $ (g @ x) Arjan _______________________________________________ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
