Use cont instead of ContT to construct.
Yes, this works.
Is this because I was using the Cont monad transformer instead of a plain coninutation monad? And with a plain
continuation monad (String -> String) -> String would have worked?
Does everybody use the transformer these days? Is a plain continuation monad still around?
Am 07/12/2014 05:12 PM, schrieb David McBride:
> It's because the type of f is not (String -> String) -> String
>
> It is (String -> Identity String) -> Identity String
>
> Do a replacement manually and you'll see that f has to be of type -> ContT String Identity String --> ContT (String ->
> Identity String) -> Identity String)
>
> You can see that in the error message Expected type: ContT String Identity String, Actual type: ContT Char [] String.
> The reason why it looks weird is that it is assuming that your monad instead of being identity is [], and that the r in
> "m r" must be a Char in order for that to work. I'm not really sure why it chose list, probably type defaulting rules.
>
>
> On Sat, Jul 12, 2014 at 6:24 AM, martin <martin.drautzburg@web.de <mailto:martin.drautzburg@web.de>> wrote:
>
> Hello all,
>
> I just started trying to understand Continuations, but my very first exercise already left me mystified.
>
> import Control.Monad.Cont
>
> resultIs :: Int -> Cont String String
> resultIs i = ContT $ f
> where
> f :: (String -> a) -> a
> f k = k ("result=" ++ show i)
>
> If resultIs returns a Cont String String, then f should be (String->String)->String, but that doesn't compile. Why is
> that so?
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