-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Apply parentheses from the right.
So: (.) :: (b -> c) -> (a -> b) -> a -> c is the same as: (.) :: (b -> c) -> (a -> b) -> (a -> c) is the same as: (.) :: (b -> c) -> ((a -> b) -> (a -> c)) How you read that is up to you, but here is one way of reading it: "accepts a function a to c and returns a function. The function returned takes a function a to b and returns a function a to c" The expression f(g(x)) in C-style languages is similar to (f . g) x Tony Morris http://tmorris.net/ PR Stanley wrote: > Hi > (.) :: (b -> c) -> (a -> b) -> (a -> c) > While I understand the purpose and the semantics of the (.) operator I'm > not sure about the above definition. > Is the definition interpreted sequentially - (.) is a fun taht takes a > fun of type (b -> c) and returns another fun of type (a -> b) etc? > Any ideas? > Thanks, Paul > > _______________________________________________ > Haskell-Cafe mailing list > [email protected] > http://www.haskell.org/mailman/listinfo/haskell-cafe > > > -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.4.6 (GNU/Linux) Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org iD8DBQFG9JzXmnpgrYe6r60RAjDrAJ0SvkZHtNsctWNYHjqxjp9lnpNvgACfS/2r 9jwUvD29/ZMMot8x3/nvyI8= =xSzA -----END PGP SIGNATURE----- _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
