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Laikoni
Laikoni
12:39 PM
@Potato44 Nice, you could post it to code review to get some feedback.
 
 
5 hours later…
Challenger5
Challenger5
5:50 PM
Is there a function like (a -> b -> c) -> (d -> a) -> (e -> b) -> d -> e -> c?
So f recip exp (*) = \x y -> recip x * exp y
@Leo I don't think that works; (<$>) for functions is just (.); (f <*> g) x is f x (g x), which might be what you meant
 
H.PWiz
H.PWiz
6:38 PM
Oct 11 at 22:09, by H.PWiz
@flawr That's liftA2 :: (a -> b -> c) -> f a -> f b -> f c where f is the function functor. This without the import goes f<$>g<*>h which is equivalent to \x -> f (g x) (h x)
@Challenger5 ^
 
Challenger5
Challenger5
@H.PWiz I've seen that; it almost works but applies both functions to the same argument. I want it to return a function that takes two arguments
 
H.PWiz
H.PWiz
Oh, I didn't read that properly...
How about ((flip.((.).)).).(.)? It's not looking good. D:
 
Challenger5
Challenger5
Oh god...
 
H.PWiz
H.PWiz
 
Challenger5
Challenger5
Yeah, I tried that too. Maybe if you reorder the arguments?
Nope, no results for any permutation of the first 3 arguments
 
H.PWiz
H.PWiz
6:50 PM
Also, didn't realise f<$>g<*>h could be just f.g<*>h. Thanks!
 

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