Michael,

A of a * a is more memory efficient than A of (a * a). In effect, a  
variant IS a tuple. If you wish to restrict yourself to restrict  
yourself to unary constructors in your programs, you're free to do so  
at the cost of extra allocations.

On 2008-05-29, at 21:13, Michael Vanier 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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— Gordon