I am learning F#. I have written custom fold function for list (with help here on StackOverflow). I am now trying to write foldback function say myOwnFoldBack. The expected output might be myOwnFoldBack (+) 0 [1; 2; 3 ] should return 6.
Here is my code for myOwnFold
let rec myOwnFoldf s0 =
function
| [] -> s0
| x::tail -> myOwnFoldf (f s0 x) tail
This works fine.
Here is code for myOwnFoldBack
let rec myOwnFoldBack f s0 =
function
| [] -> 0
| x::tail -> x f (myOwnFoldBackf tail)
The error I get is:
Type mismatch. Expecting a
'a
but given a
('b -> 'a -> int) list -> int
The resulting type would be infinite when unifying ''a' and '('b -> 'a -> int) list -> int'
I think I figured it out finally!
let rec myOwnFoldBack f s0 =
function
| [] -> s0
| x::tail -> f x (myOwnFoldBack f s0 tail)
Related
I am trying to define a function that is similar to Lisp's apply. Here is my attempt:
type t =
| Str of string
| Int of int
let rec apply f args =
match args with
| (Str s)::xs -> apply (f s) xs
| (Int i)::xs -> apply (f i) xs
| [] -> f
(* Example 1 *)
let total = apply (fun x y z -> x + y + z)
[Int 1; Int 2; Int 3]
(* Example 2 *)
let () = apply (fun name age ->
Printf.printf "Name: %s\n" name;
Printf.printf "Age: %i\n" age)
[Str "Bob"; Int 99]
However, this fails to compile. The compiler gives this error message:
File "./myprog.ml", line 7, characters 25-30:
7 | | (Str s)::xs -> apply (f s) xs
^^^^^
Error: This expression has type 'a but an expression was expected of type
string -> 'a
The type variable 'a occurs inside string -> 'a
What is the meaning of this error message? How can I fix the problem and implement apply?
You cannot mix an untyped DSL for data:
type t =
| Int of int
| Float of float
and a shallow embedding (using OCaml functions as functions inside the DSL) for functions in apply
let rec apply f args =
match args with
| (Str s)::xs -> apply (f s) xs (* f is int -> 'a *)
| (Int i)::xs -> apply (f i) xs (* f is string -> 'a *)
| [] -> f (* f is 'a *)
The typechecker is complaining that if f has type 'a, f s cannot also have for type 'a since it would mean that f has simultaneously type string -> 'a and 'a (without using the recursive types flag).
And more generally, your function apply doesn't use f with a coherent type: sometimes it has type 'a, sometimes it has type int -> 'a, other times it would rather have type string -> 'a. In other words, it is not possible to write a type for apply
val apply: ??? (* (int|string) -> ... *) -> t list -> ???
You have to choose your poison.
Either go with a fully untyped DSL which contains functions, that can be applied:
type t =
| Int of int
| Float of float
| Fun of (t -> t)
exception Type_error
let rec apply f l = match f, l with
| x, [] -> f
| Fun f, a :: q -> apply (f a) q
| (Int _|Float _), _ :: _ -> raise Type_error
or use OCaml type system and define a well-typed list of arguments with a GADT:
type ('a,'b) t =
| Nil: ('a,'a) t
| Cons: 'a * ('b,'r) t -> ('a -> 'b,'r) t
let rec apply: type f r. f -> (f,r) t -> r = fun f l ->
match l with
| Nil -> f
| Cons (x,l) -> apply (f x) l
EDIT:
Using the GADT solution is quite direct since we are using usual OCaml type without much wrapping:
let three = apply (+) (Cons(1, Cons(2,Nil)))
(and we could use a heterogeneous list syntactic sugar to make this form even lighter syntactically)
The untyped DSL requires to build first a function in the DSL:
let plus = Fun(function
| Float _ | Fun _ -> raise Type_error
| Int x -> Fun(function
| Float _ | Fun _ -> raise Type_error
| Int y -> Int (x+y)
)
)
but once we have built the function, it is relatively straightforward:
let three = apply_dsl plus [Int 2; Int 1]
type t =
| Str of string
| Int of int
| Unit
let rec apply f args =
match args with
| x::xs -> apply (f x) xs
| [] -> f Unit
Let's go step by step:
line 1: apply : 'a -> 'b -> 'c (we don't know the types of f, args and apply's return type
line 2 and beginning of line 3: args : t list so apply : 'a -> t list -> 'c
rest of line 3: Since f s (s : string), f : string -> 'a but f t : f because apply (f s). This means that f contains f in its type, this is a buggy behaviour
It's actually buggy to call f on s and i because this means that f can take a string or an int, the compiler will not allow it.
And lastly, if args is empty, you return f so the return type of f is the type of f itself, another buggy part of this code.
Looking at your examples, a simple solution would be:
type t = Str of string | Int of int
let rec apply f acc args =
match args with x :: xs -> apply f (f acc x) xs | [] -> acc
(* Example 1 *)
let total =
apply
(fun acc x ->
match x with Int d -> d + acc | Str _ -> failwith "Type error")
0 [ Int 1; Int 2; Int 3 ]
(* Example 2 *)
let () =
apply
(fun () -> function
| Str name -> Printf.printf "Name: %s\n" name
| Int age -> Printf.printf "Age: %i\n" age)
() [ Str "Bob"; Int 99 ]
Since you know the type you want to work on, you don't need GADT shenanigans, just let f handle the pattern matching and work with an accumulator
I want to implement the List.assoc function using List.find, this is what I have tried:
let rec assoc lista x = match lista with
| [] -> raise Not_found
| (a,b)::l -> try (List.find (fun x -> a = x) lista)
b
with Not_found -> assoc l x;;
but it gives me this error:
This expression has type ('a * 'b) list but an expression was expected of type 'a list
The type variable 'a occurs inside 'a * 'b
I don't know if this is something expected to happen or if I'm doing something wrong. I also tried this as an alternative:
let assoc lista x = match lista with
| [] -> raise Not_found
| (a,b)::l -> match List.split lista with
| (l1,l2) -> let ind = find l1 (List.find (fun s -> compare a x = 0))
in List.nth l2 ind;;
where find is a function that returns the index of the element requested:
let rec find lst x =
match lst with
| [] -> raise Not_found
| h :: t -> if x = h then 0 else 1 + find t x;;
with this code the problem is that the function should have type ('a * 'b) list -> 'a -> 'b, but instead it's (('a list -> 'a) * 'b) list -> ('a list -> 'a) -> 'b, so when I try
assoc [(1,a);(2,b);(3,c)] 2;;
I get:
This expression has type int but an expression was expected of type
'a list -> 'a (refering to the first element of the pair inside the list)
I don't understand why I don't get the expected function type.
First off, a quick suggestion on making your assoc function more idiomatic OCaml: have it take the list as the last argument.
Secondly, why are you attempting to implement this in terms of find? It's much easier without.
let rec assoc x lista =
match lista with
| [] -> raise Not_found
| (a, b) :: xs -> if a = x then b else assoc x xs
Something like this is simpler and substantially more efficient with the way lists work in OCaml.
Having the list as the last argument, even means we can write this more tersely.
let rec assoc x =
function
| [] -> raise Not_found
| (a, b) :: xs -> if a = x then b else assoc x xs
As to your question, OCaml infers the types of functions from how they're used.
find l1 (List.find (fun s -> compare a x = 0))
We know l1 is an int list. So we must be trying to find it in an int list list. So:
List.find (fun s -> compare a x = 0)
Must return an int list list. It's a mess. Try rethinking your function and you'll end up with something much easier to reason about.
So I have this exercise:
filter (fun x -> x = 0) [(1,0);(2,1);(3,0);(4,1)];;
result int list [1;3]
So basically you have to match your x in fun with the second number in list and if its the same you create new list with the first number.
My solution but is wrong
let rec filter f = function
| []->[]
| x::l -> if f=snd x then fst x :: filter f l else [];;
I get the following error when i want to try the code:
Error: This expression has type int but an expression was expected of
type
int -> bool
I can't reproduce the problem you report. Here's what I see when I try your code:
$ ocaml
OCaml version 4.02.1
# let rec filter f = function
| []->[]
| x::l -> if f=snd x then fst x :: filter f l else [] ;;
val filter : 'a -> ('b * 'a) list -> 'b list = <fun>
# filter 0 [(1,0); (2,1); (3,0)];;
- : int list = [1]
There are no errors, but it gets the wrong answer. That's what I would expect looking at your code.
The error that you are getting is saying that somewhere the compiler is expecting an int -> bool function, but you are giving it an int. The reason you get this error is because you have an equality (f = snd x), where f is of type int -> bool and snd x is of type int. both arguments given to the equality must be of the same type. Instead, what you want to do is simply branch on the result of applying f to the second element of x, such as:
let rec filter f = function
| []->[]
| x::l -> if f (snd x) then fst x :: filter f l else [];;
That said, I would recommend using pattern matching instead of fst and snd, such as:
let rec filter f l =
match l with
| [] -> []
| (x,y)::l -> if f y then x :: filter f l else filter f l
Note that f y will return something of type bool, which will then determine which branch to take.
Altough Matts answer is right. It's good to just reuse existing functions instead of writing a special from the ground up:
[(1,0);(2,1);(3,0);(4,1)]
|> List.filter (fun (_, x) -> x = 0)
|> List.map fst
I saw in the OCaml manual this example to use GADT for an AST with function application:
type _ term =
| Int : int -> int term
| Add : (int -> int -> int) term
| App : ('b -> 'a) term * 'b term -> 'a term
let rec eval : type a. a term -> a = function
| Int n -> n
| Add -> (fun x y -> x+y)
| App(f,x) -> (eval f) (eval x)
Is this the right way of representing function application for a language not supporting partial application?
Is there a way to make a GADT supporting function application with an arbitrary number of arguments?
Finally, is GADT a good way to represent a typed AST? Is there any alternative?
Well, partial eval already works here:
# eval (App(App(Add, Int 3),Int 4)) ;;
- : int = 7
# eval (App(Add, Int 3)) ;;
- : int -> int = <fun>
# eval (App(Add, Int 3)) 4 ;;
- : int = 7
What you don't have in this small gadt is abstraction (lambdas), but it's definitely possible to add it.
If you are interested in the topic, there is an abundant (academic) literature. This paper presents various encoding that supports partial evaluation.
There are also non-Gadt solutions, as shown in this paper.
In general, GADT are a very interesting way to represent evaluators. They tend to fall a bit short when you try to transform the AST for compilations (but there are ways).
Also, you have to keep in mind that you are encoding the type system of the language you are defining in your host language, which means that you need an encoding of the type feature you want. Sometimes, it's tricky.
Edit: A way to have a GADT not supporting partial eval is to have a special value type not containing functions and a "functional value" type with functions. Taking the simplest representation of the first paper, we can modify it that way:
type _ v =
| Int : int -> int v
| String : string -> string v
and _ vf =
| Base : 'a v -> ('a v) vf
| Fun : ('a vf -> 'b vf) -> ('a -> 'b) vf
and _ t =
| Val : 'a vf -> 'a t
| Lam : ('a vf -> 'b t) -> ('a -> 'b) t
| App : ('a -> 'b) t * 'a t -> 'b t
let get_val : type a . a v -> a = function
| Int i -> i
| String s -> s
let rec reduce : type a . a t -> a vf = function
| Val x -> x
| Lam f -> Fun (fun x -> reduce (f x))
| App (f, x) -> let Fun f = reduce f in f (reduce x)
let eval t =
let Base v = reduce t in get_val v
(* Perfectly defined expressions. *)
let f = Lam (fun x -> Lam (fun y -> Val x))
let t = App (f, Val (Base (Int 3)))
(* We can reduce t to a functional value. *)
let x = reduce t
(* But we can't eval it, it's a type error. *)
let y = eval t
(* HOF are authorized. *)
let app = Lam (fun f -> Lam (fun y -> App(Val f, Val y)))
You can make that arbitrarly more complicated, following your needs, the important property is that the 'a v type can't produce functions.
I really don't understand what's going on. I have the following code:
let rec interleave n l =
match l with
[] -> [[n]]
| head::tail -> (n::l)::(List.map (~f:fun y -> head::y) (interleave n tail))
in let rec aux l =
match l with
[] -> [l]
| head::tail -> List.concat ( List.map (interleave head) (aux tail) )
When compiling with ocaml it compiles and works as expected but under corebuild it gives me the following error:
The expression has type 'a list -> `a list list but an expression was
expected of type 'b list
Does it have something to do with labels again (as you see from ~f:fun y -> ... it has already annoyed me before)? If yes what kind of label should I use and where?
You need to reread part of manual about labeled arguments.
let rec interleave n l =
match l with
| [] -> [[n]]
| head::tail -> (n::l)::(List.map ~f:(fun y -> head::y) (interleave n tail))
and aux l =
match l with
| [] -> [l]
| head::tail -> List.concat ( List.map ~f:(interleave head) (aux tail) );;
N.B. right syntax for labeled arguments is ~label:expression
N.B. In Core List.map function has type 'a list -> f:('a -> 'b) -> 'b list and if you forget to add label to your function f it will try to unify 2nd argument with a function. That's why you have so weird error message.