On Tue, 07 Nov 2000, Chris Hecker wrote:
 
> How does the compiler "know" the datastructure is purely functional?

It doesn't. Why would it have to?

> In other words, is there any way for me to get a type into a tree that has a
> mutable member (perhaps through an abstract type),

Yes there is. You can make a map that contains, for example,  
values of type `int ref' or `int array'.

> and if so, what happens in
> that case?

Well, as you already might know, O'Caml *never* copies anything unless
you explicitely tell it to. For example, the following session:

# let ar1 = [|1|];;
val ar1 : int arr# ar1.(0) <- # ar2;;
- : int array = [|2|]2;;
- : unit = ()  ay = [|1|]
# let ar2 = ar1;;
val ar2 : int array = [|1|]
# ar2;;
- : int array = [|1|] 
# ar1.(0) <- 2;;
- : unit = ()  
# ar2;;
- : int array = [|2|]

So I mutated ar1, but the update is reflected in ar2. In C++ speech, all
variables are implicitely "references" to the underlying object. ar1 and ar2
refer to the same object. If you want ar2 to be a new array, then you
have to use Array.copy to explicitely tell O'Caml that you want a new
array.

So yes, if I have a map map1, and I get an "updated" map2 from that, then
mutations to the objects held in map1 will be reflected in map2.

Stephan
-- 
ir. Stephan H.M.J. Houben
tel. +31-40-2474358 / +31-40-2743497
e-mail: stephanh@win.tue.nl