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Simple idea for making a function infix
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Date: -- (:)
From: Till Varoquaux <till.varoquaux@g...>
Subject: Re: [Caml-list] Simple idea for making a function infix
Hi,

I don't really understand the point of the */ operator i your
definition. It's fuy ecause I've cosiderig the same problem recetly ad
came dow to this approach ( I'm redefining(@)... oe c )

On 11/13/06, Keisuke Nakano <ksk@mist.i.u-tokyo.ac.jp> wrote:
> Hi all,
>
> Haskell people sometimes complain about that OCaml cannot make an
> arbitrary function infix. For example, they can write (3 `min` 4)
> to get the result of (min 3 4) in Haskell. Can we satisfy them
> without changing OCaml's syntax?
>
> Here is a simple idea for making a function infix in OCaml.
> I hope it will be useful for those who like Haskell's backquote
> notation `function_name`.
> The idea doesn't require any change of OCaml's syntax.
>
> We use the following two infix operators.
>
>    let ( /* ) x y = y x
>    and ( */ ) x y = x y
>
> Then we can make an infix operator /*f*/ for a binary function f.
> For example, using binary functions 'min' and 'max', we can write
>
>    3 /*min*/ 4 + 6 /*max*/ 8
>
> to get 11 as 'min 3 4 + max 5 8'. Note that the infix operator
> ( */ ) may conflict with Num.( */ ) if the Num module is loaded
> and opened. You can use other definitions in a similar manner, though.
>
> You have to take care of the precedence. For example,
>
>    3 /*min*/ 4 * 6 /*max*/ 8
>
> will return 18 as 'max ((min 3 4) * 6) 8'. So we should write
>
>    (3 /*min*/ 4) * (6 /*max*/ 8)
>
> to get 24 as 'min 3 4 * max 6 8'.
>
>
> The original idea was introduced in my blog a few months ago
> (written in Japanese, though). At that time, I used other definitions:
>
>    let ( <| ) x y = y x
>    and ( |> ) x y = x y
>
> or
>
>    let ( @^ ) x y = y x
>    and ( ^@ ) x y z = x z y
>
> where the definition of ( ^@ ) should be given in a different way
> because of the precedence of infix operaters starting with '^' or '@'.
> These operators perform a different behavior because of the precedences
> of operators.
>
>    3 <|min|> 4 + 6 <|max|> 8 (* = max (min 3 (4 + 6)) 8 => 8 *)
>    3 @^min^@ 4 + 6 @^max^@ 8 (* = min 3 (max (4 + 6) 8) => 3 *)
>
> So you have to write
>
>    (3 <|min|> 4) + (6 <|max|> 8)
>    (3 @^min^@ 4) + (6 @^max^@ 8)
>
> to get 11 as min 3 4 + max 5 8'.
>
>
> Sincerely,
>
> -----------------------------------------------------------------------
> Keisuke Nakano
> Department of Mathematical Informatics,
> University of Tokyo
>
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Hi,

  I don't really understand the point of (*/) in your solution (maybe
for aesthetic
  reasons?). Actually, I tried something similar very recently (and
was talking about it a couple of minutes before reading your mail):

  ( * might interact badly with camlp4 *)
let ($) f a= a f;;

let plus a b= a+b;;

4 $plus 4;;