OCaml 中类型 ('a -> 'b) list -> 'a -> 'b list 的函数

Question

Write any Ocaml function whose type is ('a -> 'b) list -> 'a -> 'b list

('a -> 'b) list is the part that confuses me the most. I'm new to OCaml and having a hard time understanding how to write a function to get a specific datatype type.

# let int x = x+1;;
# let fcn = [int; int];;

So I'm passing a function a function and a variable. I'm going to take that variable an add it to each element of the list and return the list?

Answer 1

('a -> 'b) means a function which goes from type 'a to type 'b . Basically you need to make a function which takes a list of functions that take 'a and return 'b , plus a specific 'a value, and which returns a list of 'b values (probably by applying each function of the list of functions to the specific 'a value).

Answer 2

As this is homework, I will not provide you with a complete solution. But, as a hint, I would suggest that you take a look at this implementation of the familiar map function:

let rec map f = function
  | [] -> []
  | x :: xs -> f x :: map f xs

It has type ('a -> 'b) -> 'a list -> 'b list which means that it takes as its first argument a function that takes values of some type 'a to values of some type 'b , as its second argument a list of elements of type 'a , and that it produces a list of elements of type 'b . It proceeds by pattern matching on the argument list and, recursively applying the function ( f ) to every element x of the list.

Now have a look at the type of the function that you have to write? What does it tell you about the required behaviour of that function? Keeping the implementation of the map function in mind, how would you write your function?

Answer 3

('a -> 'b) list -> 'a -> 'b list

This means that your function has two parameters

• A list of ('a -> 'b) which represents a function taking an element of type 'a as a parameter and returning an element of type 'b . As you can see, these types are abstract, so they could be of any types for instance (int -> int) or (int -> float) etc...
• An elements of types 'a . Notice that this type must be the same as the parameter of your function.

So you'll build the resulting list with the element you give as a parameter.

Here is a little example:

let action l a =
  let rec todo l res =
    match l with
      | [] -> res
      | h :: t -> todo t res@[h a] in
  todo l []

so here, any function of type int -> int will be accepted. The same thing goes for any other type as long as you don't mix them with other types.

Answer 4

let rec func f a = match f with          (* ( 'a->'b ) list -> 'a -> 'b list *)
                       |[]->[]
                       |x::lr -> x a :: func lr a;; 

that may help ! it works fine
1 - So as we know , ocaml create the type of our function line by line

2 - in this function we have two arguments f and a

3 - ( 'a->'b ) list : for f

4 - 'a : for a
! how ocaml did that ? listen !

5 - when we matched f with [ ] 'blank list' ocaml release that is a list (****)list but doesn't know what contains yet in the last line of the code he will do ok ? nice !
- here we are in the last line of the code and we have only f of type list -

6 - x :: lr means we sort the first element of the element that is matched before : f and we add a here ocaml gives a type for a and for the list elements which is matched : f as first elements ocaml gives them types from 'a to 'z so here we have ('a->'b) list for f and 'a for a
-here we have f of type : ('a->'b) list , and a of type : 'a

7 - the result of this function 'b list so it's up to you to answer in comment ! :D thank you