michael rice nowgate at yahoo.com
Sat Jul 31 14:24:08 EDT 2010

```>From Hoogle:

Query: (:[])

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

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 at benmachine.co.uk> wrote:

From: Ben Millwood <haskell at benmachine.co.uk>
To: "michael rice" <nowgate at yahoo.com>
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?"
>
> 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.

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