haskell - Understanding how Either is an instance of Functor -

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in free time i'm learning haskell, beginner question.

in readings came across example illustrating how either a made instance of functor:

instance functor (either a)     fmap f (right x) = right (f x)     fmap f (left x) = left x

now, i'm trying understand why implementation maps in case of right value constructor, doesn't in case of left?

here understanding:

first let me rewrite above instance as

instance functor (either a)     fmap g (right x) = right (g x)     fmap g (left x) = left x

now:

  1. i know fmap :: (c -> d) -> f c -> f d

  2. if substitute f either a fmap :: (c -> d) -> either c -> either d

  3. the type of right (g x) either (g x), , type of g x d, have type of right (g x) either d, expect fmap (see 2. above)

  4. now, if @ left (g x) can use same reasoning type either (g x) b, either d b, not expect fmap (see 2. above): d should second parameter, not first! can't map on left.

is reasoning correct?

this right. there quite important reason behavior: can think of either b computation, may succeed , return b or fail error message a. (this also, how monad instance works). it's natural, functor instance won't touch left values, since want map on computation, if fails, there's nothing manipulate.


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