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