Re: [Haskell-cafe] Question on leazy evaluation
Still on the same subject, I solved one problem but don't really understand why it didn't work in the first place ! Although it seems to be a difference between newtype and data ! I have a newtype defined as : data Fct s a = Fct (s - [a]) and a function defined by : plus :: Fct a b - Fct a b - Fct a b plus (Fct f1) (Fct f2) = Fct ( \ a - (f1 a) ++ (f2 a) ) For some reason, this function does not use leazy evaluation ! I can test it using : test_fct :: Fct Int Int test_fct = Fct( \ i - [i] ) value = head $ apply (foldr plus test_fct $ repeat test_fct) 12 ... trying to get value does not terminate ! But if I change the type declaration into : newtype Fct s a = Fct (s - [a]) ... it works ! The plus function uses leazy evaluation and value can be computed. Now, I didn't really understood the difference between newtype and data. Can anyone explain that behaviour ? Thank you, Pierre -- Pierre Barbier de Reuille INRA - UMR Cirad/Inra/Cnrs/Univ.MontpellierII AMAP Botanique et Bio-informatique de l'Architecture des Plantes TA40/PSII, Boulevard de la Lironde 34398 MONTPELLIER CEDEX 5, France tel : (33) 4 67 61 65 77fax : (33) 4 67 61 56 68 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Question on leazy evaluation
On Fri, Mar 25, 2005 at 03:13:48PM +0100, Pierre Barbier de Reuille wrote: plus :: Fct a b - Fct a b - Fct a b plus (Fct f1) (Fct f2) = Fct ( \ a - (f1 a) ++ (f2 a) ) For some reason, this function does not use leazy evaluation ! I can test it using : test_fct :: Fct Int Int test_fct = Fct( \ i - [i] ) value = head $ apply (foldr plus test_fct $ repeat test_fct) 12 ... trying to get value does not terminate ! That's because you pattern match on (Fct f2) in plus. If Fct is defined with 'data', this causes the second argument of plus to be evaluated. But if I change the type declaration into : newtype Fct s a = Fct (s - [a]) ... it works ! The plus function uses leazy evaluation and value can be computed. If Fct is defined with 'newtype', it doesn't cause evaluation, because Fct is a kind of virtual / non-existent / zero-cost data constructor, so there is nothing to evaluate. Try these definitions: plus (Fct f1) ~(Fct f2) = Fct ( \ a - (f1 a) ++ (f2 a) ) plus (Fct f1) f2 = Fct ( \ a - (f1 a) ++ (apply f2 a) ) The first uses an irrefutable pattern, the second doesn't pattern match on Fct in the second argument. Best regards Tomasz ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
