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2021-04-05 02:44:50 ezrakilty joins (~ezrakilty@97-126-95-37.tukw.qwest.net)
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2021-04-05 02:58:42 <ukari> I found a interesting thing that `foo . bar1 <*> bar2 <*> bar3` can be rewrite in the form of `foo <$> bar1 <*> bar2 <*> bar3`, anyone explain this? https://gist.github.com/ukari/1175b994394f4e8526a905e9eb08ee30
2021-04-05 03:00:20 <hololeap> ukari: it's because ((->) a) has a Functor instance which corresponds to (.)
2021-04-05 03:01:57 <ukari> oh
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2021-04-05 03:02:04 <hololeap> :t not . not
2021-04-05 03:02:05 <lambdabot> Bool -> Bool
2021-04-05 03:02:10 <hololeap> :t fmap not not
2021-04-05 03:02:11 <lambdabot> Bool -> Bool
2021-04-05 03:05:08 <hololeap> don't use the Functor instance of ((->) a). it leads to confusing code.
2021-04-05 03:05:25 × stree quits (~stree@68.36.8.116) (Ping timeout: 265 seconds)
2021-04-05 03:05:42 <hololeap> if you absolutely have to, use (Reader a) instead. that has a nice name that everyone understands
2021-04-05 03:06:06 × merijn quits (~merijn@83-160-49-249.ip.xs4all.nl) (Ping timeout: 240 seconds)
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2021-04-05 03:07:27 <ukari> hololeap, is this code `foo <$> bar1 <*> bar2 <*> bar3` a example of using the Functor instance of ((->) a)? I saw it everywhere
2021-04-05 03:08:11 <hololeap> it could be. ((->) a) has an Applicative and Monad instance as well. `foo <$> bar1 <*> bar2 <*> bar3` would work for _any_ Applicative
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2021-04-05 03:09:57 <ukari> thanks you for explain this
2021-04-05 03:10:24 <hololeap> the people that write code like that for ((->) a) would be discredting the Republic of Haskell and could be summarily executed
2021-04-05 03:12:20 <ukari> :)
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2021-04-05 03:17:42 <hololeap> :t (\f g h -> (f <$> g <*> h)) :: (x -> y -> z) -> (a -> x) -> (a -> y) -> a -> z
2021-04-05 03:17:43 <lambdabot> (x -> y -> z) -> (a -> x) -> (a -> y) -> a -> z
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2021-04-05 03:18:25 <hololeap> that's what heresy looks like
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2021-04-05 04:43:36 <wroathe> Is there a way to get this to work? Also, why isn't it working? https://gist.github.com/JustinChristensen/bcffe268fa71664e820db62dd2a47a37
2021-04-05 04:44:06 <wroathe> I've tried using KindSignatures to tell it that t is a unary type constructor, but that doesn't appear to work
2021-04-05 04:44:18 <shachaf> It's not working for the reason it says.
2021-04-05 04:44:41 <shachaf> Show instances must have kind *. You can't have "instance Show Maybe", only "instance Show (Maybe a)".
2021-04-05 04:45:17 <wroathe> shachaf: Right, so how would I then write this?
2021-04-05 04:48:47 <shachaf> Hmm, you could write instance (Show a, Show (t (Bar a))) => Show (Foo t a), but that's an UndecidableInstance and probably not what you want.
2021-04-05 04:48:50 <wroathe> shachaf: I thought the solution might have been Foo (t a) a, but that's not right either
2021-04-05 04:48:55 <shachaf> There's a class Show1 for this.
2021-04-05 04:49:10 <shachaf> I guess that's the answer. https://hackage.haskell.org/package/base-4.15.0.0/docs/Data-Functor-Classes.html#t:Show1
2021-04-05 04:49:26 <wroathe> Interesting. Thanks!
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