sml - What does function return when "function times zero" in functional programming? -

i stuck sml assignment. trying create compound function (fun compound n f). it's supposed apply function f on n times example, compound 3 f equal f(f(f(x))). got work except case n zero. asked professor won't tell me direct answer. tried give me hint "what's function times zero?" still can't figure out either. can stackoverflow figure out?

thanks.

my code:

fun compound n f =     if n < 2         if n = 0 fn x => f x else fn x => f x     else fn x => f(compound (n-1) f(x)); 

example:

val fnc = fn x => x + 1; (* example function used *) compound 5 fnc(10); (* return 15 correct*) compound 0 fnc(10); (* returns 11, should 10 *) 

answer:

fun compound n f =     if n < 2         if n = 0 fn x => x else fn x => f x     else fn x => f(compound (n-1) f(x)); 

i won't give final answer because don't upset teachers ;) however, i'll try derivation believe you'll find easy complete.

let's start simple case. let's "reimplement" function application, i.e., let's write function takes function , argument , apply first param second one:

fun apply f = f 

let's use contrived function, increments integers, testing:

- fun inc n = n + 1; val inc = fn : int -> int  - inc 1; val = 2 : int  - apply inc 1; val = 2 : int 

now, let's write apply2, function takes function , argument , applies param function 2 times argument:

fun apply2 f = f (f a) 

let's test inc:

- apply2 inc 1; val = 3 : int 

seems working. might expect, we'd implement apply3, apply4 , on. let's see of them @ once:

fun apply f = f fun apply2 f = f (f a) fun apply3 f = f (f (f a)) fun apply4 f = f (f (f (f a))) 

it looks can rewrite later ones in terms of earlier ones:

fun apply2 f = f (apply f a) fun apply3 f = f (apply2 f a) fun apply4 f = f (apply3 f a) 

we can rewrite apply:

fun apply f = f (apply0 f a) 

remember previous definition of apply, they're equivalent:

fun apply f = f 

so, should apply0 be?

fun apply0 f = ...