[Haskell-beginners] fmap Maybe

Oliver Charles ollie at ocharles.org.uk
Mon Apr 27 10:13:16 UTC 2015


Unfortunately you are running in to strange behavior due to a lack of
parenthesis. First of all, let's see what your original expression actually
is:

fmap (\x -> x) Just 2 = (fmap (\x -> x) Just) 2

So you can see that you are actually fmaping over Just, rather than Just 2.
What does this mean? Well, let's ask GHC:

:t fmap _ Just
    Found hole `_' with type: Maybe a -> b

That is, when you try and fmap over the *function* Just, we have to provide
a function that expects a Maybe a, rather than an a. In your first case,
your providing the identity function, which fits the required type as Maybe
a -> Maybe a. However, in your second example, you are trying to provide
apply the (+ 2) function to a Maybe a value (because x :: Maybe a). You
cannot in general add numbers to Maybe a, hence the error message.

Your final expression works because $ is effectively providing parenthesis:

fmap (\x -> x + 2) $ Just 2 = fmap (\x -> x + 2) (Just 2)

The precedence of application in Haskell can be confusing at the start - I
suggest liberally using parenthesis, and then using HLint to remove the
ones that you don't need. Eventually, you'll build up the necessary
intuition.

Hope this helps,
- Ollie

On Mon, Apr 27, 2015 at 11:02 AM, Shishir Srivastava <
shishir.srivastava at gmail.com> wrote:

> Hi,
>
> Please can someone explain why these two expression behave differently -
>
> ----
> fmap (\x -> x) Just 2
> Just 2
>
> -----
> fmap (\x -> x+2) Just 2
>
> *No instance for (Num (Maybe a0)) arising from a use of `it'*
> *In a stmt of an interactive GHCi command: print it*
>
> ----
> The first expression evaluates fine whereas the second one fails. However
> if I do -
> ----
>
> fmap (\x -> x+2) $ Just 2
> Just 4
> ----
>
> Then the second expression also returns the Maybe value. Why is $ needed
> in second expression but not in the first one ?
>
> Thanks,
> Shishir
>
>
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