syntax bug: copying records

syntax bug: copying records

[skaller]

If x is a record then { x } should be a copy of it. But this
doesn't work, I am using this ugly workaround instead:

{ x with dummy_field = x.dummy_field }

Surely there is a better way? 

Oo.copy doesn't work either. Yes, yes, imperative programming
sucks but I'm using the standard Lexing stuff.

-- 
John Skaller <skaller at users dot sf dot net>
Felix, successor to C++: http://felix.sf.net

Re: [Caml-list] syntax bug: copying records

[Mackenzie Straight]

On 12/20/05, skaller <skaller@users.sourceforge.net> wrote:
> If x is a record then { x } should be a copy of it. But this
> doesn't work, I am using this ugly workaround instead:
>
> { x with dummy_field = x.dummy_field }
>
> Surely there is a better way?

Perhaps:

let copy (x:'a) = (Obj.obj (Obj.dup (Obj.repr x)) : 'a)

Re: [Caml-list] syntax bug: copying records

[Jacques Garrigue]

From: Mackenzie Straight <eizneckam@gmail.com>
> On 12/20/05, skaller <skaller@users.sourceforge.net> wrote:
> > If x is a record then { x } should be a copy of it. But this
> > doesn't work, I am using this ugly workaround instead:
> >
> > { x with dummy_field = x.dummy_field }
> >
> > Surely there is a better way?
> 
> Perhaps:
> 
> let copy (x:'a) = (Obj.obj (Obj.dup (Obj.repr x)) : 'a)

Not a very good idea:

  # copy 3;;
  Segmentation fault

This also explains why you need at least one record field in the
"with" notation, otherwise there is no way to know for sure the type
of the record you're copying.

Jacques Garrigue

Re: [Caml-list] syntax bug: copying records

[skaller]

On Wed, 2005-12-21 at 19:41 +0900, Jacques Garrigue wrote:
> From: Mackenzie Straight <eizneckam@gmail.com>
> > On 12/20/05, skaller <skaller@users.sourceforge.net> wrote:
> > > If x is a record then { x } should be a copy of it. But this
> > > doesn't work, I am using this ugly workaround instead:
> > >
> > > { x with dummy_field = x.dummy_field }
> > >
> > > Surely there is a better way?

> This also explains why you need at least one record field in the
> "with" notation, otherwise there is no way to know for sure the type
> of the record you're copying.

Why not?

If I write

	let x = expr in ...

then the type of x is determined by the type of expr unified
with the uses of x, right?

So the same inference doesn't apply to

	let x = { expr } in ..

instead, { .. } is required to have a definite type?
Couldn't this be checked instead, so that one could write:

	let x = { expr : thetype } in

which IMHO is preferable to picking one of the fields,
since the latter is fragile wrt the names of the fields.

How would you copy Lexing.lexbuf .. whose contents 
could be construed as implementation dependent?

Or worse .. a private record type .. which you *cannot*
copy with functional update .. although perhaps one could
argue if there is no copy constructor given it shouldn't
be copyable .. hmm.

-- 
John Skaller <skaller at users dot sf dot net>
Felix, successor to C++: http://felix.sf.net

Re: [Caml-list] syntax bug: copying records

[Mackenzie Straight]

On 12/21/05, Jacques Garrigue <garrigue@math.nagoya-u.ac.jp> wrote:
>   # copy 3;;
>   Segmentation fault

I know. You can check the tag first if you really want. Like this:

let copy x =
  let r = Obj.repr x in
  if Obj.is_block r then
    Obj.obj (Obj.dup r)
  else
    x

Of course, we all know that using Obj is equivalent to using a dirty
syringe you found on the sidewalk, so... Danger!

> This also explains why you need at least one record field in the
> "with" notation, otherwise there is no way to know for sure the type
> of the record you're copying.

True. Well, you don't actually need to know the type (indeed, the
runtime system needs to have enough information to copy values). But I
imagine it could have disastrous consequences when copying abstract
values anyway.