2

I can multiply a list per 2:

(* 2) <$> [1, 2, 3]

But I want multiply the elements which are Just:

(* 2) <$> [Just 1, Nothing, Just 3]

Error:

* Non type-variable argument in the constraint: Num (Maybe a) (Use FlexibleContexts to permit this) * When checking the inferred type it :: forall a. (Num (Maybe a), Num a) => [Maybe a] Prelude Data.List

Another try:

fmap (* 2) [Just 1, Nothing, Just 3]

Error:

* Non type-variable argument in the constraint: Num (Maybe a) (Use FlexibleContexts to permit this) * When checking the inferred type it :: forall a. (Num (Maybe a), Num a) => [Maybe a]

I have tried more things: map2, fmap2, map (*2) map, etc.

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3 Answers 3

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10

A straightforward solution is adding another fmap to get through the Maybe layer:

GHCi> fmap (* 2) <$> [Just 1, Nothing, Just 3] [Just 2,Nothing,Just 6]

Alternatively, that can be expressed using Compose, which allows handling the two functor layers as one:

GHCi> import Data.Functor.Compose GHCi> (* 2) <$> Compose [Just 1, Nothing, Just 3] Compose [Just 2,Nothing,Just 6] GHCi> getCompose $ (* 2) <$> Compose [Just 1, Nothing, Just 3] [Just 2,Nothing,Just 6]
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7

You need to map twice: once to get inside the list, second time to get inside the maybe. The operator <$> only maps once, and you can't use the same operator twice, so you have to add a call to fmap:

fmap (* 2) <$> [Just 1, Nothing, Just 3]
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1 Comment

Sure you can, the syntax is ((* 2) <$>) <$> [Just 1, Nothing, Just 3]. Only do this if you like being deliberately obtuse.
2

You can compose 2 fmap functions:

(fmap . fmap) (*2) [Just 1, Nothing, Just 3]

The first will be the fmap of the list, the second is the fmap of the maybe.

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