Re: [Caml-list] syntax question

[Adam Granicz <granicz.adam@vnet.hu>]

Hi Mike,

I suspect this was a choice made by the compiler writers - by  
special-casing constructors (thus making them different from ordinary  
functions) you avoid some "complications" (and create new ones), for  
instance no partial constructors (although IMHO there is a valid case for  
them - instead you have to resort to eta expansion) - this forces that  
each constructor fulfills its "arity".

AFAIK, the revised syntax takes care of the above issues and you can  
construct testme values as Foo 1 2.

Regards,
Adam.

On Fri, 30 May 2008 03:13:09 +0200, Michael Vanier  
<mvanier@cs.caltech.edu> wrote:

> Adam,
>
> I realize that this is how it works, but I don't understand why it  
> should work this way. AFAIK elsewhere in ocaml "int * int" always refers  
> to a tuple.  Similarly, if testme's Foo really took two int arguments I  
> would expect to be able to create Foos as "Foo 1 2" instead of "Foo (1,  
> 2)" which looks like Foo takes a single tuple argument, not two int  
> arguments.  I don't see why "int * int" and "(int * int)" are different  
> things.
>
> Mike
>
> Adam Granicz wrote:
>> Hi Michael,
>>  In the type definition
>>
>>> # type testme = Foo of int * int;;
>>  the constructor Foo takes *two* int arguments (thus, you can not  
>> construct a testme value supplying only one argument), whereas in
>>
>>> # type testme2 = Foo2 of (int * int);;
>>  it takes *one* tuple argument.
>>  Regards,
>> Adam.
>>  On Fri, 30 May 2008 00:23:40 +0200, Michael Vanier  
>> <mvanier@cs.caltech.edu> wrote:
>>
>>> Hi everyone,
>>>
>>> I got bitten by a simple syntax problem:
>>>
>>> # let a = (1, 2);;
>>> val a : int * int = (1, 2)
>>> # type testme = Foo of int * int;;
>>> type testme = Foo of int * int
>>> # Foo a;;
>>> The constructor Foo expects 2 argument(s),
>>> but is here applied to 1 argument(s)
>>> # Foo (1, 2);;
>>> - : testme = Foo (1, 2)
>>> # type testme2 = Foo2 of (int * int);;
>>> type testme2 = Foo2 of (int * int)
>>> # Foo2 a;;
>>> - : testme2 = Foo2 (1, 2)
>>>
>>> Why does the compiler treat int * int and (int * int) in type  
>>> definitions so differently?  Is it to give clearer error messages in  
>>> the typical case?
>>>
>>> Mike
>>>
>>>
>>>
>>>
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