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@afsalthaj
Created Jun 29, 2018
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Forgetted Invariant Functors !


Invariant Functors

People tend to forget this quite often. An invariant functor or an exponential functor is:

https://gist.github.com/77923fc3b7583cf7bfc6c2b5b92fba21


Covariant Functor

That's the famous Functor ! Covariant functor implements xmap by discarding g: B => A

https://gist.github.com/e7f0a2910c403d0cf04fcc6bb29de583


Contravariant Functor

As expected, it discards f: A => B and makes use of contramap to implement xmap

https://gist.github.com/f91e1ea7e6801dc91f9b328e6a12367b


Example for Contravariant Functor

https://gist.github.com/b658245bf47f647f02513213ce6aa719


Usage

https://gist.github.com/67259a39c54dc7ae897a35882ff94288


Example for covariant Functor

As expected, it is DecodeJson, where the type parameter in the type class comes at covariant position (method result)

https://gist.github.com/334fe6c8841721540bb719c8cdffd66f


Note

If type parameters are at covariant position, that means the method return contains the type.

If type parameters are at contravariant position, that means the method parameters contain the type.


When is invariant functor?

We may have types at covariant (output) or contravariant (input) position. However, we may sometime deal with both covariance and contravariance in the same type class.

Let's bring in EncodeJson and DecodeJson into one type class.


EncodeJson and DecodeJson

https://gist.github.com/d5320b66c64ea82580495476d33a76cd


Functor but invariant

So an individual map or contramap to upcast (or downcast) an A to B in the context of F[_] is not possible if F has types both in covariant and contravariant positions. It means, F has to have an invariant functor for it!


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