Re: [Haskell-cafe] beginner's problem about lists

11 Oct 2006


      Malcolm Wallace wrote:
...
Matthias Fischmann  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

Re: [Haskell-cafe] beginner's problem about lists

David Sabel