OCaml中的递归：与Concat反向

Question

I have a question regarding the execution flow in recursive functions in OCaml. This is scenario:

I have two recursive functions, Concat and Reverse . Reverse calls Concat. Would anyone be able to explain what happens when I, for example, submits the list [1; 2; 3]?

let rec concat (l1,l2) =
match l1 with
[] -> l2
| (h::t) -> h::(concat (t,l2));;

let rec reverse (l: int list) =
match l with
[] -> []
| (h :: t) -> concat (reverse t, [h]);;

// Call
let list1 = [1; 2; 3] ;;
reverse list1 ;;

I know this is not the optimal way to reverse a list, but right now I only interested in how two recursive functions work with each other.

Thanks!

Answer 1

If you annotate concat as taking two lists of ints:

let rec concat (l1, l2 : int list * int list) =
. . .

You can ask the toplevel (OCaml REPL) to trace the function calls and return values. This might tell you exactly what you want to know.

$ rlwrap ocaml
        OCaml version 4.06.1
. . .
# trace concat;;
concat is now traced.
# trace reverse;;
reverse is now traced.

# reverse [1; 2; 3];;
reverse <-- [1; 2; 3]
reverse <-- [2; 3]
reverse <-- [3]
reverse <-- []
reverse --> []
concat <-- ([], [3])
concat --> [3]
reverse --> [3]
concat <-- ([3], [2])
concat <-- ([], [2])
concat --> [2]
concat --> [3; 2]
reverse --> [3; 2]
concat <-- ([3; 2], [1])
concat <-- ([2], [1])
concat <-- ([], [1])
concat --> [1]
concat --> [2; 1]
concat --> [3; 2; 1]
reverse --> [3; 2; 1]
- : int list = [3; 2; 1]

Answer 2

Functions, recursive or not, are evaluated by first evaluating all it's arguments in an unspecified order and then calling the function with them.

So for example concat (reverse t, [h]) will first evaluate reverse and then call concat.