Re: Another rule guestion.
Simon Peyton-Jones wrote: The way it is now, ONE rule, the fold/build rule deals with fusion. The others are all there to express things in terms of fold and build, AND THEN to return certain common patterns of usage for foldr to their familiar mapList and filterList forms. Does that help? Tremendously. I was thinking that those rules were there becuase the rule "map f (map g xs) = map (f.g) xs" would simply not work. Now back to my orignal question: Why doesn't my "replaceMany (map f l) a = replaceManyMap f l" rule--as shown in the previous email titled "Rule Question"--behave as expected. -- Kevin Atkinson [EMAIL PROTECTED] http://metalab.unc.edu/kevina/
RE: Another rule guestion.
For someone who couldn't make head of tail of it, you are spot on. That's just the way it works. You ask why not have a rule for map f (map g xs) = map (f.g) xs Because then we'd need one for map f (filter g xs) and one for map f (take n xs) and so on and so on and so on. The way it is now, ONE rule, the fold/build rule deals with fusion. The others are all there to express things in terms of fold and build, AND THEN to return certain common patterns of usage for foldr to their familiar mapList and filterList forms. Does that help? Simon -Original Message- From: Kevin Atkinson Sent: Tuesday, June 29, 1999 8:34 PM To: [EMAIL PROTECTED] Subject: Another rule guestion. I was looking through PrelBase.lhs and I can not make head or tails out of the point in all the run around with build, mapFB, etc. is. As I see it the simplification of "(map f. map g) l" would go as follows: (map f . map g) xs map f (map g xs) build$ \c n - foldr (mapFB c f) n (map g xs) build$ \c n - foldr (mapFB c f) n (build$ \c n - foldr (mapFB c g) n xs) --with rule: foldr k z (build g) = g k z build$ \c n - (\c n - foldr (mapFB c g) n xs) (mapFB c f) n build$ \c n - foldr (mapFB (mapFB c f) g) n xs -- with rule: mapFB (mapFB c f) g = mapFB c (f.g) build$ \c n - foldr (mapFB c (f.g)) n xs (\c n - foldr (mapFB c (f.g)) n xs) (:) [] foldr (mapFB (:) (f.g)) [] xs -- with rule: foldr (mapFB (:) f) [] = mapList f mapList (f.g) xs which seams like a hell of a lot of work when the rule "map f (map g xs) = map (f.g) xs" should do the trick. I am sure all this run-around is here for a good reason, but what is it? I am also sure that understanding what all this run-around does is the key to understanding why the rule "replaceMany (map f l) a = replaceManyMap f l" is not behaving as expected. Thanks in advance for any help with this. I can see how the rule rewriting mechanism can be very powerful; however, it is not very useful if I can't figure out how it works... -- Kevin Atkinson [EMAIL PROTECTED] http://metalab.unc.edu/kevina/
Another rule guestion.
I was looking through PrelBase.lhs and I can not make head or tails out of the point in all the run around with build, mapFB, etc. is. As I see it the simplification of "(map f. map g) l" would go as follows: (map f . map g) xs map f (map g xs) build$ \c n - foldr (mapFB c f) n (map g xs) build$ \c n - foldr (mapFB c f) n (build$ \c n - foldr (mapFB c g) n xs) --with rule: foldr k z (build g) = g k z build$ \c n - (\c n - foldr (mapFB c g) n xs) (mapFB c f) n build$ \c n - foldr (mapFB (mapFB c f) g) n xs -- with rule: mapFB (mapFB c f) g = mapFB c (f.g) build$ \c n - foldr (mapFB c (f.g)) n xs (\c n - foldr (mapFB c (f.g)) n xs) (:) [] foldr (mapFB (:) (f.g)) [] xs -- with rule: foldr (mapFB (:) f) [] = mapList f mapList (f.g) xs which seams like a hell of a lot of work when the rule "map f (map g xs) = map (f.g) xs" should do the trick. I am sure all this run-around is here for a good reason, but what is it? I am also sure that understanding what all this run-around does is the key to understanding why the rule "replaceMany (map f l) a = replaceManyMap f l" is not behaving as expected. Thanks in advance for any help with this. I can see how the rule rewriting mechanism can be very powerful; however, it is not very useful if I can't figure out how it works... -- Kevin Atkinson [EMAIL PROTECTED] http://metalab.unc.edu/kevina/
