Please implement the function:
composeApplicative :: (Applicative f) => f (b -> c) -> f (a -> b) -> f (a -> c)
Such that:
(composeApplicative f g) <*> x == f <*> (g <*> x)
Or alternatively, explain why this can not be done?
1 Answer
It can be done:
composeApplicative p q = (.) <$> p <*> q
For more information, read the documentation for Applicative functors, more specifically, the composition law. It is effectively a statement that any Applicative instance, composeApplicative f g <*> x must always be equal to f <*> (g <*> x).
As a minor technical note, when doing equational reasoning, the left- and right-hand sides of equations must be separated with a single equals sign (=). The double-equals sign (==) is reserved for decidable runtime equality checks.
answered Jul 12, 2015 at 16:32
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"As a minor technical note, when doing equational reasoning, the left- and right-hand sides of equations must be separated with a single equals sign (=). The double-equals sign (==) is reserved for decidable runtime equality checks." This is not a consensus rule; plenty of high-profile Haskellers use
==for equational reasoning. Not even the GHC documentation follows it consistently; see e.g. the docs forFunctorandMonad. Commented Jul 13, 2015 at 18:31 -
@LuisCasillas: Point taken. But I think it's useful to distinguish between equality as a proposition (which may or may not be decidable), and decision procedure for equality (which in general doesn't exist for arbitrary types). Commented Jul 13, 2015 at 19:11
composeApplicative p q = (.) <$> p <*> q