LazyList.map2 is not as lazy as LazyList.map

[mars0i]

When an element in the result of LazyList.map2 is accessed, this forces the next element to be constructed. LazyList.map doesn't have this behavior:

# let pr = Printf.printf "%d\n%!";;
val pr : int -> unit = <fun>
# let s1 = LL.seq 0 (fun n -> let n' = n + 1 in pr n'; n') (fun _ -> true);;
val s1 : int LL.t = <lazy>
# let s2 = LL.seq 0 (fun n -> let n' = n + 1 in pr n'; n') (fun _ -> true);;
val s2 : int LL.t = <lazy>
# let s3 = LL.seq 0 (fun n -> let n' = n + 1 in pr n'; n') (fun _ -> true);;
val s3 : int LL.t = <lazy>
# let s1' = LL.map (fun n -> n)  s1;;
val s1' : int LL.t = <lazy>
# let s2s3 = LL.map2 (fun x y -> x, y) s2 s3;;
val s2s3 : (int * int) LL.t = <lazy>
# LL.hd s1';;
- : int = 0
# LL.hd s2s3;;
1
1
- : int * int = (0, 0)
# LL.hd (LL.tl s1');;
1
- : int = 1