Running a consumer inside a pipe

Question

I need to compose a consumer and a pipe so that the output of the consumer would feed the input of the pipe.

I guess this could be solved with a combinator like this:

Consumer i m r -> (r -> Producer o m r') -> Pipe i o m r'

or this:

Consumer i m i' -> Pipe i' o m r -> Pipe i o m r

or a lifting function like the following:

Consumer i m r -> Pipe i o m r

or like this:

Consumer i m o -> Pipe i o m r

I tried doing consumer >~ pipe without a success. So how to approach this?

Do you intend the consumer to run multiple times? – danidiaz Jul 14 '14 at 17:57 — danidiaz, Jul 14 '14 at 17:57
No. I don't intend that. – Nikita Volkov Jul 14 '14 at 18:03 — Nikita Volkov, Jul 14 '14 at 18:03

score 5 · Answer 1 · answered Jul 14 '14 at 21:13

For your first type signature, if your Consumer and Producer use the polymorphic type synonyms Consumer' and Producer', then the combinator you want is (>>=):

(>>=) :: Pipe i o m r -> (r -> Pipe i o m r') -> Pipe i o m r'

A Consumer' i m r will type-check as a Pipe i o m r. Similarly, a Producer' o m r' will type-check as a Pipe i o m r'.

danidiaz · Accepted Answer · 2014-07-14T19:09:04.143

Something similar to your signature Consumer i m o -> Pipe i o m r could be done like this:

{-# LANGUAGE RankNTypes #-}
import Pipes

foo :: Monad m => Consumer' i m o -> Pipe i o m ()
foo consumer = consumer >>= yield

I have used the Consumer' polymorphic type synonym which, since it is not really closed "downstream", it can be used as a Pipe that never actually yields. To make it yield the return value of the consumer, we simply use a monadic bind.

As for your Consumer i m r -> Pipe i o m r signature, it's simply the identity using the polymorphic type synonym:

iden :: Monad m => Consumer' i m r -> Pipe i o m r
iden consumer = consumer

Running a consumer inside a pipe

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