Re: [Caml-list] failwith, raise and type inference

[Ker Lutyn <ker527mail@yahoo.com>]

I wonder if we could have Haskell's $ operator added to OCaml. It would make
this idiom and a lot of my code nicer.

    failwith $ "foo" ^ "bar"

Defining it myself as

    let ($) f x = f x

doesn't work because of precedence and associativity. The caml sources
frequently define

    let (++) x f = f x

which is a similar idea - but lots of times Haskell's $ is what you want.



--- Pierre Weis <pierre.weis@inria.fr> wrote:
> [...]
> > Admitedly caml syntax is not very  beginner friendly.
> > But the observed parsing has some internal logics.
> > 
> > On usualy understand that  f x + y is in fact  (f x) + y
> > But here, with failwith being a ``special'' function and ^ a ``special''
> > operator, well...
> 
> That could be part of the problem: failwith is not ``special'' nor is ^,
> hence the regular treatment of f x ^ y as (f x) ^ y and failwith x ^ y
> as (failwith x) ^ y.
> 
> Operator precedence in Caml has been carefully crafted, you should
> have a look at the programming guide lines that gives some hints on the
> internal logics of some parts of the parsing, in the section
> 
>    When to use parentheses within an expression
> 
> Best regards,
> 
> Pierre Weis
> 
> INRIA, Projet Cristal, Pierre.Weis@inria.fr, http://pauillac.inria.fr/~weis/
> 
> 
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