On 26/04/07, skaller <skaller@users.sourceforge.net> wrote:

It knows the type of the function expression, and that is all
that is required. Incidentally Ocaml evaluates right to left. So

        f x y z

will be roughly:

        push (eval z)
        push (eval y)
        push (eval x)
        push (eval f)
        apply
        apply
        apply

But that doesn't explain how does each apply know what to do, either to build a new closure (in the case above, the first two applies) or to actually call the code (the third apply).

- Tom