Re: [Caml-list] what do I need to know to understand camlp4

[David House <dmhouse@gmail.com>]

On 24 September 2010 01:15, Elias Gabriel Amaral da Silva
<tolkiendili@gmail.com> wrote:
> [1] Pervasives should define it. In fact, even though ** is
> right-associative, it looks like any user-defined operator is
> left-associative by default. So it works like Haskell:
>
> # let ($) a b = a b;;
> val ( $ ) : ('a -> 'b) -> 'a -> 'b = <fun>
> # fun x y z -> x $ y $ z;;
> - : ('a -> 'b -> 'c) -> 'a -> 'b -> 'c = <fun>
> # let q = fun x y z -> x $ y $ z;;
> val q : ('a -> 'b -> 'c) -> 'a -> 'b -> 'c = <fun>
> # let q' = fun x y z -> (x $ y) $ z;;
> val q' : ('a -> 'b -> 'c) -> 'a -> 'b -> 'c = <fun>
> # let q'' = fun x y z -> x $ (y $ z);;
> val q'' : ('a -> 'b) -> ('c -> 'a) -> 'c -> 'b = <fun>
> # let q''' = fun x y z -> x $ y z;;
> val q''' : ('a -> 'b) -> ('c -> 'a) -> 'c -> 'b = <fun>

Incidentally, ($) in Haskell is right-associative; however the
consensus in the Haskell community (in my experience) is that this is
a mistake. If it were left-associative, you would lose the ability to
say f $ g $ x, but this can be written f . g $ x anyway (dot is
function composition (a -> b) -> (b -> c) -> a -> c, and does
right-associate), but many things would require fewer parentheses,
e.g. f (g x) (h y) can be written f $ g x $ h y.

In fact, the strict function application operator ($!) *is* left-associative.

See http://hackage.haskell.org/trac/haskell-prime/wiki/ChangeDollarAssociativity
for more information.

Also, you say "Pervasives should define it" -- the important thing
about dollar in Haskell is that it has very low precedence, hence its
ability to save parentheses. I didn't think OCaml allowed us to
specify the operator precedence of the infix operators we define.