[Haskell-beginners] How can I lift my function taking a function to a Monad?

[Simon Peter Nicholls]

On Thu, 10 Dec, 2015 at 6:43 PM, Kim-Ee Yeoh <ky3 at atamo.com> wrote:
> The wrong way is to try to write:
> 
> neededLift :: (Monad m) => ((b -> b) -> a -> a) -> (m b -> m b) -> m 
> a -> m a
> 
> which, as Daniel pointed out, is impossible.
> 
> So we start with the isomorphisms:
> 
> isoL :: (f -> a) -> (a -> f -> f) -> ((a -> a) -> (f -> f))
> isoL g s m f = s (m (g f)) f
> isoR :: (f -> a) -> ((a -> a) -> (f -> f)) -> (a -> f -> f)
> isoR _ m a = m (const a)
> 
> It turns out we really only need isoL, but isoR is there for 
> completeness.
> 
> Then we write:
> 
> liftALabel :: Applicative f => a :-> b -> f a :-> f b
> liftALabel l0 = lens g1 m1  where
>    g1 = fmap (get l0)
>    s1 = liftA2 (set l0)
>    m1 = isoL g1 s1
> 
> Naturally, monad has been effect-reduced to applicative to admit more 
> legal programs.
> 
> -- Kim-Ee

Thanks guys, I appreciate your help. You've given me a lot of food for 
thought here, Kim-Ee, and I'll spin off a new project to explore this 
in isolation.

Si
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