[Haskell-beginners] My Continuation doesn't typecheck
martin
martin.drautzburg at web.de
Sun Aug 7 11:05:02 UTC 2016
David,
I used your method of hardcoding some of the parameters to find the correct type of 'combine'. It is not at all what I
expected or wanted, but here it is:
combine :: Int -> (Int -> (Integer -> (String -> r) -> r) -> (String -> r) -> r -> String) ->
(Integer -> (String -> r) -> r) -> (String -> r) -> r -> String
Not sure what this is trying to tell me.
Am 08/06/2016 um 07:01 PM schrieb David McBride:
> 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 <mailto: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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