From Hoogle:


Query: (:[])

Error:
unexpected ":"
expecting "#", ",", "forall", "(", "[", "!" or ")"
Bad symbol



Prelude> let h = length . (:[]) . head
Prelude> h undefined
1
Prelude> :t (:[])
(:[]) :: a -> [a]
Prelude> h []
1           <======== this comes as a surprise
Prelude>

Are you saying:

[ head x ]  ->  [ *thunk* ]   and   length [ *thunk* ] ->  1, independent of what *thunk* is, even head [], i.e., *thunk* never needs be evaluated?

Michael




--- On Sat, 7/31/10, Ben Millwood <haskell@benmachine.co.uk> wrote:

From: Ben Millwood <haskell@benmachine.co.uk>
Subject: Re: [Haskell-cafe] Laziness question
To: "michael rice" <nowgate@yahoo.com>
Cc: haskell-cafe@haskell.org
Date: Saturday, July 31, 2010, 1:47 PM

On Sat, Jul 31, 2010 at 5:59 PM, michael rice <nowgate@yahoo.com> wrote:
>
> OK, in f, *length* already knows it's argument is a list.
>
> In g, *length* doesn't know what's inside the parens, extra evaluation there. So g is already ahead before we get to what's inside the [] and ().

According to the types, we already know both are lists. The question
is, of course, what kind of list.

> But since both still have eval x to *thunk* : *thunk*,  g evaluates "to a deeper level?"
>
> Michael
>

I think this question is being quite sneaky. The use of head and tail
is pretty much irrelevant. Try the pointfree versions:

f = length . (:[]) . head
g = length . tail

and see if that helps you see why f is lazier than g.