[Caml-list] use identity for default function

[Caml-list] use identity for default function

[Ashish Agarwal]

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

[Jeff Meister]

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

[Gerd Stolpmann]

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?
> >
> >
> 
> --
> Caml-list mailing list.  Subscription management and archives:
> https://sympa.inria.fr/sympa/arc/caml-list
> Beginner's list: http://groups.yahoo.com/group/ocaml_beginners
> Bug reports: http://caml.inria.fr/bin/caml-bugs



-- 
------------------------------------------------------------
Gerd Stolpmann, Darmstadt, Germany    gerd@gerd-stolpmann.de
Creator of GODI and camlcity.org.
Contact details:        http://www.camlcity.org/contact.html
Company homepage:       http://www.gerd-stolpmann.de
------------------------------------------------------------

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

[Gabriel Kerneis]

On Thu, Feb 07, 2013 at 11:53:25AM -0800, Jeff Meister wrote:
> Can't you simply define it this way?
> # external id : 'a -> 'b = "%identity";;
> # let foo ?(f = id) x = f x;;

If using unsafe features is okay, you might do even shorter:

# let foo ?(f = Obj.magic) x = f x ;;
val foo : ?f:('a -> 'b) -> 'a -> 'b = <fun>

But remember that “Obj.magic is not part of the language”!
For some good reason:

# let x : int list = foo 0;;
val x : int list = []

-- 
Gabriel

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

[Goswin von Brederlow]

On Thu, Feb 07, 2013 at 11:53:25AM -0800, Jeff Meister wrote:
> 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"

That is wrong. %identity is of type 'a -> 'a. You are lying to the
type system and all matter of errors and segfaults will ensure (see
later mails).

MfG
	Goswin

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

[Jeff Meister]

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Indeed, reading this thread now I have no idea why I thought that was a
good idea in the first place!


On Sat, Feb 9, 2013 at 5:13 AM, Goswin von Brederlow <goswin-v-b@web.de>wrote:

> On Thu, Feb 07, 2013 at 11:53:25AM -0800, Jeff Meister wrote:
> > 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"
>
> That is wrong. %identity is of type 'a -> 'a. You are lying to the
> type system and all matter of errors and segfaults will ensure (see
> later mails).
>
> MfG
>         Goswin
>
> --
> Caml-list mailing list.  Subscription management and archives:
> https://sympa.inria.fr/sympa/arc/caml-list
> Beginner's list: http://groups.yahoo.com/group/ocaml_beginners
> Bug reports: http://caml.inria.fr/bin/caml-bugs
>

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

[Martin Jambon]

On 02/07/2013 11:22 AM, Ashish Agarwal 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?

What's wrong with the following:

let foo ~f x = ...
let fooi ?(f = fun x -> x) x = foo ~f x


Martin

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

[Gabriel Kerneis]

On Thu, Feb 07, 2013 at 11:59:21AM -0800, Martin Jambon wrote:
> On 02/07/2013 11:22 AM, Ashish Agarwal wrote:
> >val foo : ?f:('a -> 'b) -> 'a -> 'b
> 
> What's wrong with the following:

It does not have the requested type (for fooi).

> let foo ~f x = ...
val foo : f:('a -> 'b) -> 'a -> 'b = <fun>

> let fooi ?(f = fun x -> x) x = foo ~f x
val fooi : ?f:('a -> 'a) -> 'a -> 'a = <fun>

-- 
Gabriel

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

[Ashish Agarwal]

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On Thu, Feb 7, 2013 at 2:59 PM, Martin Jambon <martin.jambon@ens-lyon.org>wrote:


> What's wrong with the following:
>
> let foo ~f x = ...
> let fooi ?(f = fun x -> x) x = foo ~f x


I'd not even define fooi. One can easily call `foo id`, so indeed this is
not a big issue. It would just be a slight convenience to have the default,
and it's mostly out of curiosity.

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

[Jacques Garrigue]

On 2013/02/08, at 4:22, 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?

This is an old request (about as old as optional arguments).
The problem is clear enough: you want a different type when no argument is passed.
Interestingly, it would be possible to do that using GADTs:

type (_,_) optional =
  | Override : 'a -> ('a,'b) optional
  | Default : ('a,'a) optional

let foo : type a b. (a -> b, a -> a) optional -> a -> b = function
  | Override f -> f
  | Default -> fun x -> x

Unfortunately, GADTs were not there when optional arguments were introduced.
Also, there is the problem that the type gets more complicated.
(But there could be some sugar to print identical types only once)

Jacques Garrigue