Re: [Haskell-beginners] My Continuation doesn't typecheck

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 mailto:martin.drautzburg@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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