Version française
Home     About     Download     Resources     Contact us    
Browse thread
Simple idea for making a function infix
[ Home ] [ Index: by date | by threads ]
[ Search: ]

[ Message by date: previous | next ] [ Message in thread: previous | next ] [ Thread: previous | next ]
Date: -- (:)
From: Keisuke Nakano <ksk@m...>
Subject: Simple idea for making a function infix
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