Folding, function composition, monads, and laziness, oh my?

Question

I am puzzled. I can write this:

import Control.Monad

main = print $ head $ (foldr (.) id [f, g]) [3]
  where f = (1:)
        g = undefined

<

Answer 1

It depends on which monad you're working with, and how its (>>=) operator is defined.

In the case of Maybe, (>>=) is strict in its first argument, as Daniel Fischer explained.

Here are some results for a handful of other monads.

> :set -XNoMonomorphismRestriction
> let foo = (const (return 42) <=< undefined <=< return) 3
> :t foo
foo :: (Num t, Monad m) => m t

Identity: Lazy.

> Control.Monad.Identity.runIdentity foo
42

IO: Strict.

> foo :: IO Integer
*** Exception: Prelude.undefined

Reader: Lazy.

> Control.Monad.Reader.runReader foo "bar"
42

Writer: Has both a lazy and a strict variant.

> Control.Monad.Writer.runWriter foo
(42,())
> Control.Monad.Writer.Strict.runWriter foo
*** Exception: Prelude.undefined

State: Has also both a strict and a lazy version.

> Control.Monad.State.runState foo "bar"
(42,"*** Exception: Prelude.undefined
> Control.Monad.State.Strict.runState foo "bar"
*** Exception: Prelude.undefined

Cont: Strict.

> Control.Monad.Cont.runCont foo id
*** Exception: Prelude.undefined