AW: [Caml-list] use identity for default function

[Gerd Stolpmann <info@gerd-stolpmann.de>]

Nope:

#  external id : 'a -> 'b = "%identity";;
external id : 'a -> 'b = "%identity"
# (id 5);;
- : 'a = <poly>
# ((id 5) : string);;
Signal -10

# external id : 'a -> 'b = "%identity";;
external id : 'a -> 'b = "%identity"
# let foo ?(f = id) x = f x;;
val foo : ?f:('a -> 'b) -> 'a -> 'b = <fun>
# foo 5;;
- : 'a = <poly>
# ((foo 5):string);;
Signal -10

I think it is not sound what Ashish wants: If f is not passed, the type  
checker needs to add a constraint about the return value. This looks  
like a dependent type.

Gerd

Am 07.02.2013 20:53:25 schrieb(en) Jeff Meister:
> Can't you simply define it this way? (I don't think I'm using any new
> features, so maybe I misunderstood your question.)
> 
>         OCaml version 4.00.1
> 
> # external id : 'a -> 'b = "%identity";;
> external id : 'a -> 'b = "%identity"
> # let foo ?(f = id) x = f x;;
> val foo : ?f:('a -> 'b) -> 'a -> 'b = <fun>
> 
> 
> On Thu, Feb 7, 2013 at 11:22 AM, Ashish Agarwal  
> <agarwal1975@gmail.com>wrote:
> 
> > I'd like to implement
> >
> > val foo : ?f:('a -> 'b) -> 'a -> 'b
> >
> > where the default for f is identity. Is there now a way to do this  
> with
> > all the new features of OCaml?
> >
> >
> 
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Gerd Stolpmann, Darmstadt, Germany    gerd@gerd-stolpmann.de
Creator of GODI and camlcity.org.
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