[Haskell-cafe] Lazy evaluation and tail-recursion

[Daniel Fischer]

On Wednesday 16 March 2011 20:02:54, Yves Parès wrote:
> > Yes, and a tail-recursive map couldn't run in constant space
> 
> Yes, I meant "if you are consuming it just once immediately".
> 

And that's what, to my knowledge, is impossible with tail recursion. A tail 
recursive map/fmap would have to traverse the entire list before it could 
return anything.

> > the above pattern [...] is better, have the recursive call as a
> > non-strict
> 
> field of a constructor.
> 
> Which pattern? Mine or Tillman's? Or both?

Yours/the Prelude's. I hadn't seen Tillmann's reply yet when I wrote mine.
In

map f (x:xs) = (:) (f x) (map f xs)

the outermost call is a call to a constructor [that is not important, it 
could be a call to any sufficiently lazy function, so that you have a 
partial result without traversing the entire list] which is lazy in both 
fields, so a partial result is returned immediately. If the element (f x) 
or the tail is not needed, it won't be evaluated at all.
If there are no other references, the (f x) can be garbage collected 
immediately after being consumed/ignored.


Tillmann:

>   data EvaluatedList a
>   
>      =  Cons a (List a)
>      
>      |  Empty
>    
>    type List a
>    
>      = () -> EvaluatedList a
>    
>    map :: (a -> b) -> (List a -> List b)
>    map f xs
>    
>      = \_ -> case xs () of
>      
>                Cons x xs  ->  Cons (f x) (\_ -> map f xs ())
>                Empty      ->  Empty
> 
> Here, the call to map is more visibly in tail position.

According to the definition of tail recursion that I know, that's not tail 
recursive.
By that, a function is tail-recursive if the recursive call (if there is 
one) is the last thing the function does, which in Haskell would translate 
to it being the outermost call.

Thus a tail recursive map would be

map some args (x:xs) = map other args' xs

, with a worker:

map f  = go []
  where
    go ys [] = reverse ys
    go ys (x:xs) = go (f x:ys) xs