2025/06/08

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2025-06-08 09:12:18 +0200 <mauke> yes
2025-06-08 09:08:10 +0200sabathan2(~sabathan@amarseille-159-1-12-107.w86-203.abo.wanadoo.fr)
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2025-06-08 09:07:04 +0200 <yabobay> so it creates f x : acc, and if i just wanted to take the first 3 elements it would just do those and acc would be left unsolved?
2025-06-08 09:04:19 +0200 <ski> and only if `f' forces its second argument, will the recursive call actually happen
2025-06-08 09:03:52 +0200 <ski> `f' is given control first, not the recursive call (giving the "previous result", for the tail, conceptually having processed from the end)
2025-06-08 09:03:46 +0200 <ski> if Haskell would be a strict language, it would (operationally) start from the end. it still "conceptually" starts from the end (just like `foldl' conceptually (and operationally) starts at the beginning). but because of non-strictness (typically implemented by laziness, call-by-need), it *operationally* will start "outermost" in `f x (foldr f z xs)' (from `foldr f z (x:xs) = f x (foldr f z xs)'), so that
2025-06-08 09:02:18 +0200sabathan2(~sabathan@amarseille-159-1-12-107.w86-203.abo.wanadoo.fr)
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2025-06-08 09:00:40 +0200caconym7(~caconym@user/caconym) caconym
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2025-06-08 08:57:42 +0200sabathan2(~sabathan@amarseille-159-1-12-107.w86-203.abo.wanadoo.fr)
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2025-06-08 08:54:02 +0200 <yabobay> hold on, nevermind. i thought foldr started at the end. oops.
2025-06-08 08:52:42 +0200sabathan2(~sabathan@amarseille-159-1-12-107.w86-203.abo.wanadoo.fr)
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2025-06-08 08:51:54 +0200 <yabobay> i'm reading 'learn you a haskell', and it demonstrates foldr by showing you this map implementation: `map' f xs = foldr (\x acc -> f x : acc) [] xs`. doesn't this fail at being lazy? if you wanna `take 3` off of a list like this for example
2025-06-08 08:50:55 +0200yabobay(~pizza@2a02:85f:fcd5:1601:a753:ff59:be6f:253f) yabobay
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2025-06-08 08:35:58 +0200euphores(~SASL_euph@user/euphores) euphores
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2025-06-08 08:09:22 +0200merijn(~merijn@host-vr.cgnat-g.v4.dfn.nl) merijn
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