How to use a Reader type encoded with continuation passing style

Question

I was under the impression that every value of type a can be described with a rank-2 polymorphic type newtype Id a = Id {runId :: forall r. (a -> r) -> r } in continuation passing style. So I derived the following type to define the Reader accordingly:

newtype Reader e a = Reader {runReader :: forall r. ((e -> a) -> r) -> r}

Then I tried to construct a value of this type and run it:

reader f = Reader (\k -> k f) -- where is f's argument?
runReader (reader id) -- what is the 2nd argument?

Provided that Reader's encoding in CPS and its type are valid, how would I use it?

chi · Accepted Answer · 2018-02-08T18:56:35.473

1

reader f = Reader (\k -> k f) -- where is f's argument?

We don't have to pass a second argument now. This is very similar to the regular data Reader e a = Reader (e->a) and reader f = Reader f which do not involve a second argument.

runReader (reader id) -- what is the 2nd argument?

The general equation is

runReader (reader f) k = k f

so,

runReader (reader f) id x = id f x = f x

You can use f = id (which acts like ask in the real reader monad), but don't confuse it with the other id which is provided as a continuation k, and which recovers the original f.

edited Feb 08 '18 at 18:56

answered Feb 08 '18 at 18:32

chi

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I think you want `id f x = f x`, not `id f x = x`. – Daniel Wagner Feb 08 '18 at 18:53
It's probably useful to define a helper `runReader' = flip runReader id` to get rid of the continuation. – Feb 08 '18 at 21:41

How to use a Reader type encoded with continuation passing style

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