[mercury-users] typeclass constraint on type definition... wrong

[Mark Anthony BROWN]

Robert Bossy writes:
> 
> Hello,
> 
> I tried the following
> 
> :- typeclass foo(T) where [].
> :- type bar(T) ---> bar(T) <= foo(T).
> 
> 
> I meant a declaration of the type bar, parametric on the type variable T =
> since T is an instance of typeclass foo.
> And it didn't work...
> Is it semantically wrong? If not how to write it?
> 

You can just leave the type variable T unconstrained in the type definition.
Any predicate that wishes to use bar(T) as an argument type should
place the constraint on it there.  Eg:

:- type bar(T) ---> bar(T).

:- pred pred_that_uses_bar(bar(T), ...) <= foo(T).
:- mode pred_that_uses_bar(in, ...) is detism.

> The error is "undefined type foo/1"... so what does
> 
> :- type foo(T) ---> foo(T).
> :- type bar(T) ---> bar(T) <= foo(T).
> 
> mean?
> 

The second declaration parses as

:- type bar(T) ---> (bar(T) <= foo(T)).

It defines the type bar(T), for all T, which has one functor '<='.
This functor takes two arguments, one of type bar(T) and the other
of type foo(T).

Cheers,
Mark
-- 
Mark Brown  (dougl at cs.mu.oz.au)       )O+   |  For Microsoft to win,
MEngSc student,                             |  the customer must lose
Dept of Computer Science, Melbourne Uni     |          -- Eric S. Raymond
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