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Today's Topics:

   1. Re:  fmap Maybe (Alexey Shmalko)
   2. Re:  fmap Maybe (Oliver Charles)


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Message: 1
Date: Mon, 27 Apr 2015 10:12:44 +0000
From: Alexey Shmalko <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
        beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] fmap Maybe
Message-ID:
        <CAFC2PC52fCfbi=h63xG=qul6jdjznt2hayc_546acfwdtxm...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"

Hi, Shishir,

It's because Haskell uses juxtaposition for argument passing. So the first
case

fmap (\x -> x) Just 2

is equal to

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

While

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

is

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

I believe you want the latter.

Basically, the first example works because ((->) r) is an instance of
Functor.

instance Functor ((->) r) where
    fmap = (.)

So basically first example is:
((\x -> x) . Just) 2

Now you should see why it behaves this way.

Have a nice day!
Alexey Shmalko
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Message: 2
Date: Mon, 27 Apr 2015 11:13:16 +0100
From: Oliver Charles <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
        beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] fmap Maybe
Message-ID:
        <cagrp5rkx_gknxrvqdxsrdvuovcnqhysx_v5hqu5cns+z2v6...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"

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 <
[email protected]> 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
>
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
>
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