# strictness question

Simon Peyton-Jones simonpj@microsoft.com
Fri, 2 Mar 2001 08:11:41 -0800

```Strange.  You don't supply a complete program, so it's hard to
test.

Nevertheless, the Haskell Report (Sect 3.12) specifies that
a let adds a single twiddle.  Thus

let (x, (y,z)) = e in b

means

let x = case e of (x,(y,z)) -> x
y = case e of (x,(y,z)) -> y
z = case e of (x,(y,z)) -> z
in b

And that is what GHC implements.  You get something different if you

let (x, ~(y,z)) = e in b

means
let x = case e of (x,_) -> x
y = case e of (_,(y,_)) -> y
etc

Adding more twiddles means less eager matching.  I don't know whether
Hugs implements this.

Simon

| -----Original Message-----
| From: S. Doaitse Swierstra [mailto:doaitse@cs.uu.nl]
| Sent: 01 March 2001 11:26
| Subject: strictness question
|
|
| I ran into a difference between GHC and Hugs. The following code:
|
| f  (P p) ~(P q)   = P (\ k -> \inp -> let (((pv, (qv, r)), m), st) =
| p (q k) inp
|                                        in  (((pv qv  , r ), m), st))
|
| runs fine with Hugs but blows up with GHC, whereas:
|
| f  (P p) ~(P q)   = P (\ k -> \inp -> let ~(~(~(pv, ~(qv, r)), m),
| st) = p (q k) inp
|                                        in  (((pv qv  , r ), m), st))
|
| runs fine with GHC too.
|
|  From the Haskell manual I understand that pattern matching
| in "let"'s
| should be done lazily, so the addition of a collection of ~'s should
| not make a difference. Am I right with  this interpretation?
|
| A possible source of this problem may be origination from the smarter
| GHC optimiser, but in that case the optimiser is not doing its work
| well.
|
|     Doaitse Swierstra
|
|
|
|
| --
| ______________________________________________________________
| ____________
| S. Doaitse Swierstra, Department of Computer Science, Utrecht
| University
|                        P.O.Box 80.089, 3508 TB UTRECHT,   the
| Netherlands
|                        Mail:  mailto:doaitse@cs.uu.nl
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