[Solved] Composing applicative functions [closed]

Question

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.

Accepted 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.