[Haskell-cafe] Laziness question

[wren ng thornton]

Brandon S Allbery KF8NH wrote:
> michael rice wrote:
>> Are you saying:
>>
>> [ head x ]  ->  [ *thunk* ]   and   length [ *thunk* ] ->  1, independent of
>> what *thunk* is, even head [], i.e., *thunk* never needs be evaluated?
> 
> Exactly.  (I was being cagey because the first response was cagey, possibly
> suspecting a homework question although it seems like an odd time for it.)
> 
> length not only does not look inside of the thunk, it *can't* look inside
> it; all it knows is that it has a list, it specifically does *not* know what
> that list can hold.  So the only thing it can do is count the number of
> "unknown somethings" in the list.

Not entirely true:

     stupidlyStrictLength :: [a] -> Integer
     stupidlyStrictLength []     = 0
     stupidlyStrictLength (x:xs) = x `seq` 1 + stupidlyStrictLength xs

Though, of course, if we actually wanted this function we should use an 
accumulator in order to avoid stack overflow when evaluating the 
(1+(1+...0)) thunk at the end.

-- 
Live well,
~wren