[Haskell-cafe] beginner's problem about lists

[David Sabel]

Malcolm Wallace wrote:
> Matthias Fischmann <fis at wiwi.hu-berlin.de> wrote:
>
>   
>>> No, your Fin type can also hold infinite values.
>>>       
>> let q = FinCons 3 q in case q of FinCons i _ -> i  ==>  _|_
>>
>> does that contradict, or did i just not understand what you are
>> saying?
>>     
>
> That may be the result in ghc, but nhc98 gives the answer 3.
>
> It is not entirely clear which implementation is correct.  The Language
> Report has little enough to say about strict components of data
> structures - a single paragraph in 4.2.1.  It defines them in terms of
> the strict application operator ($!), thus ultimately in terms of seq,
> and as far as I can see, nhc98 is perfectly compliant here.
>
> The definition of seq is
>     seq _|_ b = _|_
>     seq  a  b = b, if a/= _|_
>
> In the circular expression
>     let q = FinCons 3 q in q
> it is clear that the second component of the FinCons constructor is not
> _|_ (it has at least a FinCons constructor), and therefore it does not
> matter what its full unfolding is.
>   
The translation given in the language report says this expression is 
equivalent to

 let q = (\x1 x2 -> (( FinCons' $ x1) $! x2)  )  3 q in q            
(where FinCons' is a "lazy" constructor)

in terms of seq

let q = (\x1 x2 -> x2 `seq`  ( FinCons'  x1 x2 ) ) 3 q in q

this evaluates to

let q = (q `seq`  ( FinCons' 3) ) in q

hence on top-level  q is rather a seq-expression than a 
constructor-expression  ;-)

Regards,
  David