I'm confused about let polymorphism in OCaml.
Consider the following code:
A:
let f = (fun v -> v) in
((f 3), (f true))
B:
let f = (fun v -> v) in
((fun g ->
let f = g in
f) f)
C:
let f = (fun v -> v) in
((fun g ->
let f = g in
((f 3), (f true))) f)
For A and B, there is no problem. But for C, OCaml reports error:
Error: This expression has type bool but an expression was expected of type
int
So for A, when evaluating ((f 3), (f true)), f's type is 'a -> 'a,
for B, when evaluating let f = g in f, f's type is 'a -> 'a.
But for C, when evaluating ((f 3), (f true)), f's type is int -> int.
Why C's f doesn't have type 'a -> 'a?
I have difficulty in understanding the implementation of OCaml's
let polymorphism, I'll appreciate it a lot if anyone can give a concise
description of it with respect to the question.
Your code is unnecessarily confusing because you're using the same name f for two different things in B and also two different things in C.
Inside C you have this function:
fun g -> let f = g in (f 3, f true)
Again this is unnecessarily complicated; it's the same as:
fun g -> (g 3, g true)
The reason this isn't allowed is that it only works if g is a polymorphic function. This requires rank 2 polymorphism, i.e., it requires the ability to define function parameters that are polymorphic.
I'm not exactly sure what you're trying to do, but you can have a record type whose field is a polymorphic function. You can then use this record type to define something like your function:
# type r = { f : 'a . 'a -> 'a };;
type r = { f : 'a. 'a -> 'a; }
# (fun { f = g } -> (g 3, g true)) { f = fun x -> x };;
- : int * bool = (3, true)
# let myfun { f = g } = (g 3, g true);;
val myfun : r -> int * bool = <fun>
# myfun { f = fun x -> x };;
- : int * bool = (3, true)
The downside is that you need to pack and unpack your polymorphic function.
As a side comment, your example doesn't seem very compelling, because the number of functions of type 'a -> 'a is quite limited.
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
for an example, if a function receives a function as a factor and iterates it twice
func x = f(f(x))
I have totally no idea of how the code should be written
You just pass the function as a value. E.g.:
let apply_twice f x = f (f x)
should do what you expect. We can try it out by testing on the command line:
utop # apply_twice ((+) 1) 100
- : int = 102
The (+) 1 term is the function that adds one to a number (you could also write it as (fun x -> 1 + x)). Also remember that a function in OCaml does not need to be evaluated with all its parameters. If you evaluate apply_twice only with the function you receive a new function that can be evaluated on a number:
utop # let add_two = apply_twice ((+) 1) ;;
val add_two : int -> int = <fun>
utop # add_two 1000;;
- : int = 1002
To provide a better understanding: In OCaml, functions are first-class
values. Just like int is a value, 'a -> 'a -> 'a is a value (I
suppose you are familiar with function signatures). So, how do you
implement a function that returns a function? Well, let's rephrase it:
As functions = values in OCaml, we could phrase your question in three
different forms:
[1] a function that returns a function
[2] a function that returns a value
[3] a value that returns a value
Note that those are all equivalent; I just changed terms.
[2] is probably the most intuitive one for you.
First, let's look at how OCaml evaluates functions (concrete example):
let sum x y = x + y
(val sum: int -> int -> int = <fun>)
f takes in two int's and returns an int (Intuitively speaking, a
functional value is a value, that can evaluate further if you provide
values). This is the reason you can do stuff like this:
let partial_sum = sum 2
(int -> int = <fun>)
let total_sum = partial_sum 3 (equivalent to: let total_sum y = 3 + y)
(int = 5)
partial_sum is a function, that takes in only one int and returns
another int. So we already provided one argument of the function,
now one is still missing, so it's still a functional value. If that is
still not clear, look into it more. (Hint: f x = x is equivalent to
f = fun x -> x) Let's come back to your question. The simplest
function, that returns a function is the function itself:
let f x = x
(val f:'a -> 'a = <fun>)
f
('a -> 'a = <fun>)
let f x = x Calling f without arguments returns f itself. Say you
wanted to concatenate two functions, so f o g, or f(g(x)):
let g x = (* do something *)
(val g: 'a -> 'b)
let f x = (* do something *)
(val f: 'a -> 'b)
let f_g f g x = f (g x)
(val f_g: ('a -> 'b) -> ('c -> 'a) -> 'c -> 'b = <fun>)
('a -> 'b): that's f, ('c -> 'a): that's g, c: that's x.
Exercise: Think about why the particular signatures have to be like that. Because let f_g f g x = f (g x) is equivalent to let f_g = fun f -> fun g -> fun x -> f (g x), and we do not provide
the argument x, we have created a function concatenation. Play around
with providing partial arguments, look at the signature, and there
will be nothing magical about functions returning functions; or:
functions returning values.
I'm encoding a form of van Laarhoven lenses in OCaml, but am having difficulty due to the value restriction.
The relevant code is as follows:
module Optic : sig
type (-'s, +'t, +'a, -'b) t
val lens : ('s -> 'a) -> ('s -> 'b -> 't) -> ('s, 't, 'a, 'b) t
val _1 : ('a * 'x, 'b * 'x, 'a, 'b) t
end = struct
type (-'s, +'t, +'a, -'b) t =
{ op : 'r . ('a -> ('b -> 'r) -> 'r) -> ('s -> ('t -> 'r) -> 'r) }
let lens get set =
let op cont this read = cont (get this) (fun b -> read (set this b))
in { op }
let _1 = let build (_, b) a = (a, b) in lens fst build
end
Here I am representing a lens as a higher order type, a transformer of CPS-transformed functions ('a -> 'b) -> ('s -> 't) (as was suggested here and discussed here). The functions lens, fst, and build all have fully generalized types but their composition lens fst build does not.
Error: Signature mismatch:
...
Values do not match:
val _1 : ('_a * '_b, '_c * '_b, '_a, '_c) t
is not included in
val _1 : ('a * 'x, 'b * 'x, 'a, 'b) t
As shown in the gist, it's perfectly possible to write _1
let _1 = { op = fun cont (a, x) read -> cont a (fun b -> read (b, x)) }
but having to manually construct these lenses each time is tedious and it would be nice to build them using higher order functions like lens.
Is there any way around the value restriction here?
The value restriction is a limitation of the OCaml type system that prevents some polymorphic values from being generalized, i.e. having a type that is universally quantified over all type variables. This is done to preserve soundness of the type system in the presence of mutable references and side effects.
In your case, the value restriction applies to the _1 value, which is defined as the result of applying the lens function to two other functions, fst and build. The lens function is polymorphic, but its result is not, because it depends on the type of the arguments it receives. Therefore, the type of _1 is not fully generalized, and it cannot be given the type signature you expect.
There are a few possible ways to work around the value restriction in this case:
Use explicit type annotations to specify the type variables you want to generalize. For example, you can write:
let _1 : type a b x. (a * x, b * x, a, b) Optic.t = lens fst (fun (_, b) a -> (a, b))
This tells the compiler that you want to generalize over the type variables a, b, and x, and that the type of _1 should be a lens that works on pairs with any types for the first and second components.
Use functors to abstract over the type variables and delay the instantiation of the lens function. For example, you can write:
module MakeLens (A : sig type t end) (B : sig type t end) (X : sig type t end) = struct
let _1 = lens fst (fun (_, b) a -> (a, b))
end
This defines a functor that takes three modules as arguments, each defining a type t, and returns a module that contains a value _1 of type (A.t * X.t, B.t * X.t, A.t, B.t) Optic.t. You can then apply this functor to different modules to get different instances of _1. For example, you can write:
module IntLens = MakeLens (struct type t = int end) (struct type t = int end) (struct type t = string end)
let _1_int = IntLens._1
This gives you a value _1_int of type (int * string, int * string, int, int) Optic.t.
Use records instead of tuples to represent the data types you want to manipulate with lenses. Records have named fields, which can be accessed and updated using the dot notation, and they are more amenable to polymorphism than tuples. For example, you can write:
type ('a, 'x) pair = { first : 'a; second : 'x }
let lens_first = lens (fun p -> p.first) (fun p b -> { p with first = b })
let lens_second = lens (fun p -> p.second) (fun p b -> { p with second = b })
This defines two lenses, lens_first and lens_second, that work on any record type that has a first and a second field, respectively. You can then use them to manipulate different kinds of records, without having to worry about the value restriction. For example, you can write:
type point = { x : int; y : int }
type person = { name : string; age : int }
let p = { x = 1; y = 2 }
let q = lens_first.op (fun x f -> x + 1) p (fun p -> p)
(* q is { x = 2; y = 2 } *)
let r = { name = "Alice"; age = 25 }
let s = lens_second.op (fun x f -> x + 1) r (fun r -> r)
(* s is { name = "Alice"; age = 26 } *)
Are the two types: int -> int -> int and int -> (int -> int) the same?
If I write let f x = fun y -> x + y + 1, utop returns int -> int -> int. But I want is a function that takes an int as a parameter and then return a function which also takes an int and returns int, i.e., int -> (int -> int)
Is there a way to do that?
Also for (’a * ’b -> ’c) -> (’a -> ’b -> ’c), I wrote let f g = fun a b -> g (a,b), but it returns (’a * ’b -> ’c) -> ’a -> ’b -> ’c, the parentheses are eliminated. But why?
If a new function is returned as result, will it be anyway curried?
Usually, in lambda-calculus all functions have 1 arguments. Functions that take two arguments (not a tuple) are functions which take 1 argument and return another functions. If you look at the problem from this point of view you will understand that -> is right-associative.
Yes, they are the same. The arrow is a right-associative infix constructor, which is why the parentheses are redundant on the right.
It is perhaps helpful to realise that
let f x y z = e
is simply syntactic sugar for
let f = fun x -> fun y -> fun z -> e
Everything else follows form there.
Well yes, they are the same.
You can try it out, just type in
let f x y = x + y + 1;;
val f : int -> int -> int = <fun>
# let g = f 1;;
val g : int -> int = <fun>
# let _ = List.map g [1;2;3];; (* would do the same with (f 1) instead of g *)
- : int list = [3; 4; 5]
The whole idea is that function are by default curryfied. And as a -> ( b -> c ) and a -> b -> c are equivalent, they are displayed as the lightest type possible.
Is there a way to change the order from left-associative to right-associative, except parentheses? For example in Haskell you can write foo $ bar b and foo will be applied to a result from bar b.
let a x = x * 4;;
let b y = y + 2;;
let c = a ??? b 3;;
print_int c;;
Should print 20
Sure, you can define it yourself:
let (###) f x = f x
Then, a ### b 3 evaluates to 20. Make sure to select a starting symbol such that it is right-associative (see here) ($... is left-associative)
You just have to define a symbol for such applications:
let (###) f x = f x ;;
And then
let f x = x * 4;;
let g y = y + 2;;
let a = f ### g 3;;
print_int a;;
does print 20.
Note that the next version of OCaml (3.13 or 4.00) will provide builtin primitives for applications that avoid creating intermediate partially applied functions:
external (###) : ('a -> 'b) -> 'a -> 'b = "%apply"
external (|>) : 'a -> ('a -> 'b) -> 'b = "%revapply"
The last one is the opposite of %apply:
print_int (3 |> g |> f);;
Note that you cannot use ($) as it is left-associative in the definition of the OCaml parser:
let ($) f x = f x ;;
let a = f $ g 3;; (* ok ! ??? *)
let a = f $ g $ g 3;; (* ERROR -> g is not an integer,
because OCaml computes (f $ g) first *)