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+5 A:

It looks like you have it right, f(g(x)) should work fine. I'm not sure why you have a return keyword there (it's not a keyword in ocaml). Here is a correct version,

let compose f g x = f (g x)

The type definition for this is,

val compose : ('b -> 'c) -> ('a -> 'b) -> 'a -> 'c = <fun>

Each, 'a,'b,'c are abstract types; we don't care what they are, they just need to be consistent in the definition (so, the domain of g must be in the range of f).

let x_plus_x_plus_1 = compose (fun x -> x + 1) (fun x -> x * 2)

nlucaroni 2010-04-29 04:14:05

Thank you very much that helped a lot! But I have another question, how is it that we are allowed to call compose without specifying a value for x?

nicotine 2010-04-29 05:59:38

Function returns function. HOF!

ygrek 2010-04-29 06:54:44

Perhaps what is happening is clearer when writing it `let compose f g = fun x -> f (g x)` but these programs are exactly the same thing (one is syntactic sugar for the other, I am not sure which). Calling `compose f g` doesn't require a value for `x` because `compose f g` doesn't use `x`: it builds and returns a function.

Pascal Cuoq 2010-04-29 08:26:34

@nicotine; Yes, it returns a functions that takes the newly defined types. In fact, this is the point of currying. So, for example, x_plus_x_plus_1 is a function of, (int -> int).

nlucaroni 2010-04-29 20:54:46

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OCaml: Currying without defined values

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