[Haskell-beginners] My Continuation doesn't typecheck

David McBride toad3k at gmail.com
Sat Aug 6 17:01:00 UTC 2016


The only way to do this is to do it step by step.
:t combine
combine :: t1 -> (t1 -> t2 -> t) -> t2 -> t

>:t combine 9
combine 9 :: Num t1 => (t1 -> t2 -> t) -> t2 -> t

>:t f1
f1 :: Int -> (Integer -> r) -> r
>:t combine 9 f1
combine 9 f1 :: (Integer -> t) -> t

>:t f2
f2 :: Integer -> (String -> r) -> r
>:t combine 9 f1 f2
combine 9 f1 f2 :: (String -> r) -> r

At some point the t2 in combine becomes a function, which causes the rest
of the type to change.  I feel like combine was meant to be something else,
f (g a) or g (f a) or something else, but I'm not sure what.


On Sat, Aug 6, 2016 at 4:03 AM, martin <martin.drautzburg at web.de> wrote:

> Hello all,
>
> in order to gain some intuition about continuations, I tried the following:
>
> -- two functions accepting a continuation
>
>         f1 :: Int -> (Integer->r) -> r
>         f1 a c = c $ fromIntegral (a+1)
>
>         f2 :: Integer -> (String -> r) -> r
>         f2 b c = c $ show b
>
>         -- combine the two functions into a single one
>
>         run1 :: Int -> (String -> r) -> r
>         run1 a = f1 a f2
>
>
>         -- *Main> run1 9 id
>         -- "10"
>
> So far so good.
>
>
> Then I tried to write a general combinator, which does not have f1 and f2
> hardcoded:
>
>         combine a f g = f a g
>
>         -- This also works
>
>         -- *Main> combine 9 f1 f2 id
>         -- "10"
>
>
> What confuses me is the the type of combine. I thought it should be
>
>         combine :: Int ->
>         (Int -> (Integer->r) -> r) ->        -- f1
>         (Integer -> (String -> r) -> r) ->   -- f2
>         ((String -> r) -> r)
>
>
> but that doesn't typecheck:
>
>         Couldn't match expected type ‘(String -> r) -> r’
>         with actual type ‘r’
>
>
> Can you tell me where I am making a mistake?
>
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