I'm just starting to play with OCaml and I wrote the following function:

let rec is_sorted lst =
    match lst with
       [] -> true
    | [elt] -> true
    | a :: b :: tail -> a<b & is_sorted(b :: tail)


I'm wondering whether the b :: tail actually allocates memory or is 
the compiler smarter than that?  What's the best way to avoid this? 
If I rewrite to match instead with "a :: tail" (and then later pull 
that apart into b :: more_tail) am I guaranteed that tail is 
non-empty because of the prior [elt] case?

-- Bruce