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
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.