Top Tweets for #ApplyingVia
The newtype unwrapping is awful and doesn't signal the intent that we want, i.e. configuring behaviour of the accumulating monoid. That's what I hoped #ApplyingVia would do but eh

@mstk @clementd @TraverseDecider @TechnoEmpress with #ApplyingVia (if that ever happens)
traverse @(via Const . Ap) (fmap singleton)
@RiccardoOdone WIP #ApplyingVia: (<=<) and (>=>) are defined directly in terms of Category composition
(<=<) = (.) @(via Kleisli)
(<=<) = (<<<) @(via Kleisli)
(>=>) = (>>>) @(via Kleisli)
@sullyj3 @alexandrionascu @ESYudkowsky And similarly
f a = f $ a
f a = a & f
we have
fmap f as = f <$> as
fmap f as = as <&> f
In #ApplyingVia notation we have another relationship
($) = (<$>) @(via Identity)
(&) = (<&>) @(via Identity)
@kmett @Blaisorblade @anormalform @NicolasRinaudo @Koisell Yes I am apparently.. shamelessly. but broadly tagless, via,.. made me appreciate *typeclasses* even more
And a ambition to explore them, for example (modulo flipping) #ApplyingVia could get us
foldr = foldMap @(via Endo)
foldl = foldMap @(via Dual . Endo)
which is my favouri.
(<>) @(via Ap) = liftA2 (<>)
with #ApplyingVia
ok! we can derive Eq and Ord for functions :) .. of the form Rep f -> a
with #ApplyingVia we can easily sort lists of functions
sort @(via FromRepresentable @Pair)
:: [Bool->Bool]
-> [Bool->Bol]
![Iceland_jack's tweet photo. ok! we can derive Eq and Ord for functions :) .. of the form Rep f -> a
with #ApplyingVia we can easily sort lists of functions
sort @(via FromRepresentable @Pair)
:: [Bool->Bool]
-> [Bool->Bol] https://t.co/qfvlTd9wRr](https://pbs.twimg.com/media/EXjOBwOX0AECG6j.jpg)
@tizusa1000 @1acson @typeswitch err
{-# Language #ApplyingVia #-}
don't :: Applicative f => f a -> f ()
don't = mempty @(via (>>> Ap))
@jedws This is my favourite application of #ApplyingVia, clear and simple.. it makes the definitions of fold{r,l} soo clear
foldr = foldMap @(via Endo)
foldl = foldMap @(via Dual . Endo)
(these hold modulo some flipping)

Don't forget to submit Haskell project ideas for Google Summer of Code. Google is literally offering to pay a student to improve your open source project full-time for 3 months
You can submit a project idea by creating a pull request here:
https://t.co/NTkXKQBwN5
On my way to be social but with #ApplyingVia we could define
sum'noNaN :: RealFloat a => [a] -> a
sum'noNaN = sum @[] @(via NoNaN)
Is this some variation of this useful

this is #ApplyingVia
total total = lmap @(via Costar) total \case
Empty -> 0
Push (a, b) -> a + b

filterM @(via Cont)
:: (a -> (Bool -> r) -> r)
-> ([a] -> ([a] -> r) -> r)
#ApplyingVia
![Iceland_jack's tweet photo. filterM @(via Cont)
:: (a -> (Bool -> r) -> r)
-> ([a] -> ([a] -> r) -> r)
#ApplyingVia https://t.co/oVUcus1bVm](https://pbs.twimg.com/media/EHYAvi5WsAAJQKj.jpg)
Boss: please come up with applying a function inside of a double continuation
fired: (. ((. ((. ((. ((. (.)) . (.))) . (.))) . (.))) . (.))) . (.)
hired: (<*>) @(via Compose . fmap cont . cont)
#ApplyingVia

@luqui It's possible (see via paper). A Representable functor f is iso to (Rep f ->), #DerivingVia/#ApplyingVia via https://t.co/8slfJKDcwq uses this iso.
fmap definition
1. map to function
2. function fmap
3. map back
toFun >>> fmap @((->) (Rep f)) f >>> fromFun
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![Iceland_jack's tweet photo. ok! we can derive Eq and Ord for functions :) .. of the form Rep f -> a
with #ApplyingVia we can easily sort lists of functions
sort @(via FromRepresentable @Pair)
:: [Bool->Bool]
-> [Bool->Bol] https://t.co/qfvlTd9wRr](https://pbs.twimg.com/media/EXjNpkGWkAA2HSl.png)
![Iceland_jack's tweet photo. ok! we can derive Eq and Ord for functions :) .. of the form Rep f -> a
with #ApplyingVia we can easily sort lists of functions
sort @(via FromRepresentable @Pair)
:: [Bool->Bool]
-> [Bool->Bol] https://t.co/qfvlTd9wRr](https://pbs.twimg.com/media/EXjNpfxXgAE3KzN.png)

![Iceland_jack's tweet photo. filterM @(via Cont)
:: (a -> (Bool -> r) -> r)
-> ([a] -> ([a] -> r) -> r)
#ApplyingVia https://t.co/oVUcus1bVm](https://pbs.twimg.com/media/EHYAviyWwAA9iGn.png)

