Strange matching-problem... ?!

Strange matching-problem... ?!

[Oliver Bandel]

Hello,

any idea about this?!





# type foobar = Foo | Bar;;
type foobar = Foo | Bar

# match Foo with
   Foo -> "foo"
  | Bar -> "bar";;
- : string = "foo"

# match Foo with
   _Foo -> "foo"
  |Bar -> "bar";;
Warning: this match case is unused.
- : string = "foo"

# match Foo with
  _ Foo -> "foo"
  | Bar -> "bar";;
Syntax error

# match Foo with
  _Bar -> "bar"
  | Foo -> "foo";;
Warning: this match case is unused.
- : string = "bar"

# type foobar2 = _Foo | Bar;;
Syntax error
# 


Why is the second match working without any error message?
Did I ovelooked something?

I found that behaviour, after I tried to
use Sumtypes starting with an "_" in the name.

Using such things like _Foo instead of Foo
does not yield any error message, but when defining
the type with _Foo then ocaml complains about my
wrong syntax.

What's going on here and why?

TIA,
   Oliver

Re: [Caml-list] Strange matching-problem... ?!

[Ville-Pertti Keinonen]

On Thu, 2005-01-27 at 13:20 +0100, Oliver Bandel wrote:

> # match Foo with
>    _Foo -> "foo"
>   |Bar -> "bar";;
> Warning: this match case is unused.
> - : string = "foo"

Here, you're matching the value Foo against a catch-all pattern with a
variable called _Foo.  This is equivalent to writing:

match <anything> with
   x -> "foo"
 | <anything> -> "bar"

The first case matches any value.

Re: [Caml-list] Strange matching-problem... ?!

[Frederic van der Plancke]

Oliver Bandel wrote:
> # match Foo with
>   _Bar -> "bar"
>   | Foo -> "foo";;
> Warning: this match case is unused.
> - : string = "bar"
> 
> # type foobar2 = _Foo | Bar;;
> Syntax error
> #
> 
> Why is the second match working without any error message?
> Did I ovelooked something?
> 
> I found that behaviour, after I tried to
> use Sumtypes starting with an "_" in the name.

Doesn't _ work as a lowercase letter ?
Then _Foo and _Bar are not valid sum type tags, but they are valid
variable names hence 

   match Foo with
   _Bar -> "bar"
   | Foo -> "foo";;

works exactly like

   match Foo with
   x -> "bar"
   | Foo -> "foo";;

would.


Frédéric