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2021-05-14 08:40:01 <boxscape> not sure if that's possible to define though
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2021-05-14 08:42:14 <boxscape> it should be possible, if this class has a `flip` method and a law that flip is an isomorphism, I guess
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2021-05-14 09:12:47 <kingswim> it seems -> can apprear in lambda \x->x,and also can apprear in fff:: a->a->a,but donot know it can do like this : data Procedure = InitBoosters (controller->script)? where i can find the origin introduction of ->?
2021-05-14 09:13:56 <boxscape> kingswim the -> in the lambda syntax is a different -> than the (->) type constructor. They're both built into the language, though ghci will say something if you ask it about the type constructor with :i
2021-05-14 09:13:59 <boxscape> % :i (->)
2021-05-14 09:14:00 <yahb> boxscape: type (->) :: * -> * -> *; type (->) = FUN 'Many :: * -> * -> *; -- Defined in `GHC.Types'; infixr -1 ->; instance Applicative ((->) r) -- Defined in `GHC.Base'; instance Functor ((->) r) -- Defined in `GHC.Base'; instance Monad ((->) r) -- Defined in `GHC.Base'; instance Monoid b => Monoid (a -> b) -- Defined in `GHC.Base'; instance Semigroup b => Semigroup (a -> b) -- Defined in `GHC.Base'; instance [s
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2021-05-14 09:15:54 <boxscape> kingswim similar to how you can have `data A = A`, and overload A to be both a type and data constructor, -> is "overloaded" to be a type constructor for functions, so `a -> b` is the type of functions from a to b, but also as part of the syntax of the function data constructor, so that `\x -> y` is a function of type `a -> b` if (x :: a) and (y ::
2021-05-14 09:15:55 <boxscape> b)
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2021-05-14 09:16:22 <boxscape> (though it's a bit different because the `->`s are built into the language, whereas `A` is not.)
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2021-05-14 09:17:19 <boxscape> in `data Procedure = InitBoosters (controller -> script)`, you're using the type constructor version, to say that the argument to InitBoosters must be a function of type controller -> script
2021-05-14 09:18:43 <kingswim> i see,i will list diff and try find out the origin reason.
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2021-05-14 09:25:44 <kuribas> why does applicativeDo not allow "do a <- x; let b = y; pure Foo x y"
2021-05-14 09:25:46 <kuribas> ?
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2021-05-14 09:26:07 <kuribas> But it allows "do a <- x; b <- pure y; pure Foo x y"
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2021-05-14 09:26:43 <kuribas> which is exactly the same
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2021-05-14 09:27:11 <kuribas> does ApplicativeDo doesn't do desugaring?
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2021-05-14 09:37:29 <boxscape> huh lambdabot's @undo doesn't support let
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2021-05-14 09:39:57 <boxscape> kuribas you can see the output with -ddump-ds, which might tell you why it doesn' work
2021-05-14 09:40:05 <enikar> kuribas: ApllicativeDo does desugaring, but with the rules of Applicative. And if you use a rule that involves Monad, then (>>=) is used, if possible.
2021-05-14 09:40:57 <kuribas> boxscape: so "let x = y;" cannot be desugared to "x <- pure y"?
2021-05-14 09:41:24 <boxscape> I think that was meant for enikar
2021-05-14 09:41:40 <kuribas> no
2021-05-14 09:41:42 <boxscape> hmm
2021-05-14 09:41:46 <boxscape> well, I don't know the answer
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2021-05-14 09:42:22 <kuribas> even so if the bindings in the let don't depend on the previous bindings, the let can be floated up, and it should still work.
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2021-05-14 09:44:36 <kuribas> "do a <- x; let b = y; pure Foo x y" => (\a -> Foo a y) <$> x
2021-05-14 09:44:40 <kuribas> looks fine to me.
2021-05-14 09:45:17 <boxscape> that does look pretty applicative
2021-05-14 09:45:31 <boxscape> functorial even
2021-05-14 09:45:59 <kuribas> or better: (let b = y in \a -> Foo a y) <$> x
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2021-05-14 09:48:08 <boxscape> (whoops lambdabot's @undo *does* support let, but the `b = y` needs to be surrounded by {})
2021-05-14 09:48:32 <kuribas> I don't see any lambdabot output
2021-05-14 09:48:48 <boxscape> I experimented in private messages
2021-05-14 09:48:54 <kuribas> @undo do a <- x; let b = y; pure $ Foo x y
2021-05-14 09:48:54 <lambdabot> <unknown>.hs:1:37:Parse error: EOF
2021-05-14 09:49:05 <kuribas> @undo do {a <- x; let b = y; pure $ Foo x y}
2021-05-14 09:49:06 <lambdabot> <unknown>.hs:1:38:Parse error: }
2021-05-14 09:49:08 <boxscape> @undo do a <- x; let {b = y}; pure $ Foo x y
2021-05-14 09:49:08 <lambdabot> x >>= \ a -> let { b = y} in pure $ Foo x y
2021-05-14 09:49:29 <boxscape> but that's without ApplicativeDo
2021-05-14 09:51:54 <kuribas> oh, this works: do a <- x; pure $ let {b = y} in Foo x y
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