[Haskell-cafe] Laziness question
nowgate at yahoo.com
Sat Jul 31 14:24:08 EDT 2010
expecting "#", ",", "forall", "(", "[", "!" or ")"
Prelude> let h = length . (:) . head
Prelude> h undefined
Prelude> :t (:)
(:) :: a -> [a]
Prelude> h 
1 <======== this comes as a surprise
Are you saying:
[ head x ] -> [ *thunk* ] and length [ *thunk* ] -> 1, independent of what *thunk* is, even head , i.e., *thunk* never needs be evaluated?
--- On Sat, 7/31/10, Ben Millwood <haskell at benmachine.co.uk> wrote:
From: Ben Millwood <haskell at benmachine.co.uk>
Subject: Re: [Haskell-cafe] Laziness question
To: "michael rice" <nowgate at yahoo.com>
Cc: haskell-cafe at haskell.org
Date: Saturday, July 31, 2010, 1:47 PM
On Sat, Jul 31, 2010 at 5:59 PM, michael rice <nowgate at 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?"
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.
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