Christopher Wilson christopher.j.wilson at gmail.com
Fri Dec 17 22:22:54 CET 2010

```On Fri, Dec 17, 2010 at 11:04 AM, michael rice <nowgate at yahoo.com> wrote:

> I don't understand this error message. Haskell appears not to understand
> that 1 is a Num.
>
> Prelude> :t 1
> 1 :: (Num t) => t
> Prelude> :t [1,2,3,4,5]
> [1,2,3,4,5] :: (Num t) => [t]
> Prelude>
>
> Michael
>
> ===================
>
> f :: [Int] -> IO [Int]
> f lst = do return lst
>
> main = do let lst = f [1,2,3,4,5]
>           fmap (+1) lst
>
> ===============================
>
> Prelude> :l test
> [1 of 1] Compiling Main             ( test.hs, interpreted )
>
> test.hs:5:17:
>     No instance for (Num [Int])
>       arising from the literal `1' at test.hs:5:17
>     Possible fix: add an instance declaration for (Num [Int])
>     In the second argument of `(+)', namely `1'
>     In the first argument of `fmap', namely `(+ 1)'
>     In the expression: fmap (+ 1) lst
> Prelude>
>
>
> _______________________________________________
>
>
Excuse any inaccuracies, I'm somewhat new at Haskell myself, but what it
looks like is happening is that at the point in main where you've bound
"lst", it will have type of "IO [Int]".  The signature for fmap is:

fmap :: (Functor f) => (a -> b) -> f a -> f b

if you call fmap (+1) the next argument that fmap "expects" is something
that is "in" just one functor, for example, this

fmap (+1) [1,2,3,4,5]

works fine, but, something that is "IO [Int]" won't.  You can compose two
'fmap's to solve this:

:t (fmap.fmap)
(fmap.fmap)
:: (Functor f, Functor f1) => (a -> b) -> f (f1 a) -> f (f1 b)

which means that 'main' looks like:

main = do let lst = f [1, 2, 3, 4, 5]
(fmap.fmap) (+1) lst

--
Chris Wilson <christopher.j.wilson at gmail.com>
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