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[Caml-list] Probably FAQ: Why is list-append (list :: elem) so expensive?
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Date: 2002-09-25 (18:39)
From: Brian Hurt <brian.hurt@q...>
Subject: Re: [Caml-list] Probably FAQ: Why is list-append (list :: elem) so expensive?

Thanks to everyone for answering my question.  I think the light is 
starting to glimmer, however dim :-).

I think I'm asking the wrong question.  Is there any circumstances where 
Ocaml will reuse the cells from one list to make another list?

Consider the following Ocaml code:

    let dup_list lst =
        let rec reverse_list l accum =
            match l with
                [] -> accum
                | head :: tail -> reverse_list tail (head :: accum)
            reverse_list (reverse_list lst)

Ok, the first time through reverse_list the compiler has to allocate a 
whole new list, including a whole new set of list cells, for the 
accumlation list.  This is because the code doesn't dare modify lst.  But 
the second time reverse_list is called, it's obvious that l is the list 
allocated the first time through, and it's garbage as soon as the call 
completes.  In fact, each cell of the list is garbage after it's been 
prepended to accum.  So why not reuse the cells from l to form the accum?  
Yes, I know this violates the immutability of l.  But it seems to me that, 
with the exception of garbage creation rates, the two are identical.

Phrased this way, I start wondering how much of an optimization this
actually would be.  Ocaml has a copying garbage collector, which means
allocating new objects is cheap.  Especially if what you're doing is
duplicating a recently allocated, soon to be garbage list- like my example

Of course, appending elements to a list would only work if it was
optimized as such a "modifiable list" in the optimizer.  Which strikes me
as a dangerous proposition.  The ability to write algorithms which are
O(N) if you compile with the right optimizations, and O(N^2) if you don't
(or if you make some subtle change to the code and all of a sudden the
optimizer can modify the list anymore) strikes me as being sub-optimal.

Heh.  One of the dangers of Ocaml is that it makes the algorithm clear 
enough that it encourages you to over-optimize your algorithms in the same 
way that C encourages you to cycle count and over-optimize your 


On Wed, 25 Sep 2002, Markus Mottl wrote:

> On Wed, 25 Sep 2002, Brian Hurt wrote:
> > I get that the cost of list concatenation is proportional to the length of 
> > the first list.  So (elem :: list) is O(1) no matter what the length of 
> > list is.  But (list :: elem) is O(n), with n being the length of the list.
> You probably mean (list @ [elem]).
> > Why doesn't Ocaml keep a pointer to the last element of the list,
> > as well as the first element?
> Because standard lists are purely functional, i.e. you can't just
> overwrite some pointer to append an element. If there are several lists
> that share a common sublist, such an operation would change both lists.
> E.g.:
>   let l = [1; 2; 3] in
>   let l1 = 0 :: l in
>   let l2 = 42 :: l in
>   l @ [4]
> If the last operation overwrote anything in "l", this would also change
> l1 and l2, thus destroying referential transparency: list variables
> couldn't be used for straightforward equational reasoning anymore,
> which is important if you want to e.g. proof list algorithms correct.
> As you noted, keeping pointers to the previous element also makes
> list nodes larger and thus more expensive both in terms of time and
> space. Why have this by default if lists are most often not used in ways
> that require efficiency with such operations?
> If you want to have mutable lists, it's not difficult to implement them.
> But be warned: you might be surprised to discover that standard lists
> most often perform better in not too extreme cases.
> In case you want to learn more about purely functional datastructures,
> you will definitely want to read Chris Okasaki's book "Purely Functional
> Data Structures", which is published by Cambridge University Press (1998).
> Regards,
> Markus Mottl

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