>     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).

The generated abstract machine code is more like:
             push (eval z)
             push (eval y)
             push (eval x)
             push (eval f)
             apply 3        (*  number of arguments provided *)

"apply" doesn't do anything clever, it just enters the code of the
called function f.  It's the code of f that determines what to do
with the arguments provided on the stack.

More details can be found in one of my talks:
http://gallium.inria.fr/~xleroy/talks/zam-kazam05.pdf

- Xavier Leroy