I am a little confused about weak polymorphism in OCaml.
Please see the following snippet, where I define a function remember:
let remember x =
let cache = ref None in
match !cache with
| Some y -> y
| None -> cache := Some x; x
;;
The compiler can infer the polymorphic type 'a -> 'a, and cache is used locally.
But when I modify the above code into
let remember =
let cache = ref None in
(fun x -> match !cache with
| Some y -> y
| None -> cache := Some x; x)
;;
the compiler infers the weakly polymorphic type '_a -> '_a, also, it seems that cache is shared across invocations of remember.
Why does the compiler infer a weakly polymorphic type here and why is cache shared?
What is more, if I change the code again
let remember x =
let cache = ref None in
(fun z -> match !cache with
| Some y -> z
| None -> cache := Some x; x)
;;
the compiler infers the polymorphic type 'a -> 'a -> 'a and cache becomes locally used. Why is this the case?
let remember =
let cache = ref None in
(fun x -> match !cache with
| Some y -> y
| None -> cache := Some x; x)
;;
Here cache is closed over by the returned function. But at the point where we declare cache, we have no information on what the type will be; it'll be determined by whatever the type of x is and cache is created when remember is defined.
But since this is a closure we can do something like this:
> remember 1
1
Now it's clear that cache : int option ref since we've actually stored something in it. Since there's only ever one cache, remember can only ever store one type.
In the next one, you close over 2 things, x and cache. Since we create a new cache ref with each invocation of remember the type can be fully polymorphic again. The reason the type isn't weakly polymorphic is because we know that we're going to store x in it and we have xs type at the time when cache is created.
This seems to do with value restriction. Full value restriction (as in SML) would reject your code altogether. Weakly polymorphic types are described in the paper "Relaxing the Value Restriction" by Jacques Garrigue, which I admittedly just stumbled upon after reading your question:
http://caml.inria.fr/pub/papers/garrigue-value_restriction-fiwflp04.pdf
The fact that cache is shared across invocations should be obvious if you have a correct mental model of what ML code means. You are defining two values, remember and cache. Nesting simply makes the scope of cache private to the block.
let remember x =
let cache = ref None in
match !cache with
| Some y -> y
| None -> cache := Some x; x
let remember x =
let cache = ref None in
(fun z -> match !cache with
| Some y -> z
| None -> cache := Some x; x)
In the above two versions, remember is a "direct" function, every time you call it like remember 1, it will initialise cache to ref None, isn't it? So actually, it does not remember anything, the cache is not shared between any remember calls.
In the other version:
let remember =
let cache = ref None in
(fun x -> match !cache with
| Some y -> y
| None -> cache := Some x; x)
it is different. remember is still a function for sure, however, the real part that defines its content is (fun x -> match ...). It includes cache and cache is initialised once and will be only once. So cache is shared between the future remember call.
Related
I'm writing a quicksort function for an exercise. I already know of the 5-line functional quicksort; but I wanted to improve the partition by having it scan through the list once and return a pair of lists splitting the original list in half. So I wrote:
fun partition nil = (nil, nil)
| partition (pivot :: rest) =
let
fun part (lst, pivot, (lesseq, greater)) =
case lst of
[] => (lesseq, greater)
| (h::t) =>
if h <= pivot then part (t, pivot, (h :: lesseq, greater))
else part (t, pivot, (lesseq, h :: greater))
in
part (rest, pivot, ([pivot], []))
end;
This partitions well enough. It gives me a signature val partition = fn : int list -> int list * int list. It runs as expected.
It's when I use the quicksort below that things start to break.
fun quicksort_2 nil = nil
| quicksort_2 lst =
let
val (lesseq, greater) = partition lst
in
quicksort_2 lesseq # quicksort_2 greater
end;
I can run the above function if I eliminate the recursive calls to quicksort_2; but if I put them back in (to actually go and sort the thing), it will cease to run. The signature will be incorrect as well, giving me val quicksort_2 = fn : int list -> 'a list. The warning I receive when I call the function on a list is:
Warning: type vars not generalized because of value restriction are instantiated to dummy types (X1,X2,...)
What is the problem here? I'm not using any ref variables; the type annotation I've tried doesn't seem to help...
The main issue is that you're lacking the singleton list base case for your quicksort function. It ought to be
fun quicksort [ ] = [ ]
| quicksort [x] = [x]
| quicksort xs =
let
val (l, r) = partition xs
in
quicksort l # quicksort r
end
which should then have type int list -> int list given the type of your partition. We have to add this case as otherwise you'll never hit a base case and instead recurse indefinitely.
For some more detail on why you saw the issues you were having though:
The signature will be incorrect as well, giving me val quicksort_2 = fn : int list -> 'a list
This is because the codomain of your function was never restricted to be less general than 'a list. Taking a look at the possible branches in your original implementation we can see that in the nil branch you return nil (of most general type 'a list) and in the recursive case you get two 'a lists (per our assumptions thus far) and append them, resulting in an 'a list---this is fine so your type is not further restricted.
[Value Restriction Warning]
What is the problem here? I'm not using any ref variables
The value restriction isn't really related to refs (though can often arise when using them). Instead it is the prohibition that anything polymorphic at the top level must be a value by its syntax (and thus precludes the possibility that a computation is behind a type abstractor at the top level). Here it is because given xs : int list we (ignoring the value restriction) have quicksort_2 xs : 'a list---which would otherwise be polymorphic, but is not a syntactic value. Correspondingly it is value restricted.
let memoize (sequence: seq<'a>) =
let cache = Dictionary()
seq {for i in sequence ->
match cache.TryGetValue i with
| true, v -> printf "cached"
| false,_ -> cache.Add(i ,i)
}
I will call my memoize function inside this function :
let isCached (input:seq<'a>) : seq<'a> = memoize input
If the given sequence item is cached it should print cached otherwise it will continue to add sequence value to cache.
Right now I have problems with types.
When I try to call my function like this :
let seq1 = seq { 1 .. 10 }
isCached seq1
It throws an error
"The type int does not match the type unit"
I want my function to work generic even though I return printfn. Is it possible to achieve that? And while adding value to the cache is it appropriate to give the same value to tuple?
eg:
| false,_ -> cache.Add(i ,i)
I think the problem is that your memoize function does not actually return the item from the source sequence as a next element of the returned sequence. Your version only adds items to the cache, but then it returns unit. You can fix that by writing:
let memoize (sequence: seq<'a>) =
let cache = Dictionary()
seq {for i in sequence do
match cache.TryGetValue i with
| true, v -> printf "cached"
| false,_ -> cache.Add(i ,i)
yield i }
I used explicit yield rather than -> because I think that makes the code more readable. With this change, the code runs as expected for me.
Tomas P beat me to the punch, but I'll post this up anyway just in case it helps.
I'm not too sure what you are trying to achieve here, but I'll say a few things that I think might help.
Firstly, the type error. Your isCached function is defined as taking a seq of type 'a, and returning a seq of type 'a. As written in your question, right now it takes a seq of type 'a, and returns a sequence of type unit. If you try modifying the output specification to seq<'b> (or actually just omitting it altogether and letting type inference do it), you should overcome the type error. This probably still won't do what you want, since you aren't actually returning the cache from that function (you can just add cache as the final line to return it). Thus, try something like:
let memoize (sequence: seq<'a>) =
let cache = Dictionary()
for i in sequence do
match cache.TryGetValue i with
| true, v -> printf "cached"
| false,_ -> cache.Add(i ,i)
cache
let isCached (input:seq<'a>) : seq<'b> = memoize input
All this being said, if you are expecting to iterate over the same sequence a lot, it might be best just to use the library function Seq.cache.
Finally, with regards to using the value as the key in the dictionary... There's nothing stopping you from doing that, but it's really fairly pointless. If you already have a value, then you shouldn't need to look it up in the dictionary. If you are just trying to memoize the sequence, then use the index of the given element as the key. Or use the specific input as the key and the output from that input as the value.
In a learning environment, what are my options to provide type signatures for functions?
Standard ML doesn't have top-level type signatures like Haskell. Here are the alternatives I have considered:
Module signatures, which require either a separate signature file, or the type signature being defined in a separate block inside the same file as the module itself. This requires the use of modules, and in any production system that would be a sane choice.
Modules may seem a little verbose in a stub file when the alternative is a single function definition. They both introduce the concept of modules, perhaps a bit early,
Using val and val rec I can have the complete type signature in one line:
val incr : int -> int =
fn i => i + 1
val rec map : ('a -> 'b) -> 'a list -> 'b list =
fn f => fn xs => case xs of
[] => []
| x::ys => f x :: map f ys
Can I have this and also use fun?
If this is possible, I can't seem to get the syntax right.
Currently the solution is to embed the argument types and the result type as such:
fun map (f : 'a -> 'b) (xs : 'a list) : 'b list =
raise Fail "'map' is not implemented"
But I have experienced that this syntax gives the novice ML programmer the impression that the solution either cannot or should not be updated to the model solution:
fun map f [] = []
| map f (x::xs) = f x :: map f xs
It seems then that the type signatures, which are supposed to aid the student, prevents them from pattern matching. I cannot say if this is because they think that the type signatures cannot be removed or if they should not be removed. It is, of course, a matter of style whether they should (and where), but the student should be enabled to explore a style of type inference.
By using a let or local bound function, and shadowing
you can declare the function, and then assign it to a value.
using local for this is more convenient, since it has the form:
local decl in decl end, rather than let decl in expr end,
meaning let's expr, wants a top-level argument f
val map = fn f => let fun map = ... in map end
I don't believe people generally use local, anymore primarily because modules can do anything that local can, and more, but perhaps it is worth considering it as an anonymous module, when you do not want to explain modules yet.
local
fun map (f : 'a -> 'b) (x::rest : 'a list) : 'b list
= f x :: map f rest
| map _ ([]) = []
in
val (map : ('a -> 'b) -> 'a list -> 'b list) = map;
end
Then when it comes time to explain modules, you can declare the structure inside the local, around all of the declarations,
and then remove the local, and try to come up with a situation, where they have coded 2 functions, and it's more appropriate to replace 2 locals, with 1 structure.
local
structure X = struct
fun id x = x
end
in val id = X.id
end
perhaps starting them off with something like the following:
exception ReplaceSorryWithYourAnswer
fun sorry () = raise ReplaceSorryWithYourAnswer
local
(* Please fill in the _'s with the arguments
and the call to sorry() with your answer *)
fun map _ _ = sorry ()
in
val map : ('a -> 'b) -> ('a list) -> ('b list) = map
end
I'm looking to add a cache to a simple compiler in OCaml. I have created a simpler version of the code that I had that reproduces the same issue. What I need for the cache is to be able to create a Map with A's as keys, so I can lookup the compile output. Here is the code:
module A
= struct
type ineq =
| LT
| EQ
| GT
type t =
...
module ACacheMapKey
= struct
type t = t
let compare a b =
match cmp a b with
| LT -> -1
| EQ -> 0
| GT -> 1
end
module CMap = Map.Make(ACacheMapKey)
let cache_map = CMap.empty
let cmp a b =
...
end
In module A, type t is a recursive AST-like type. cmp a b returns a ineq. The compile function was left out for brevity, but it just uses the cache before running through a computationally expensive process. In order to create a cache map in A, I need a compatible key module. My attempt at that is ACacheMapKey, but the type t = t doesn't refer to the parent. The error it gives is Error: The type abbreviation t is cyclic. So, is there a better way to make a cache over A? Or is there an easy way to reference the parent and make my current structure work?
Type definitions, unlike let bindings, are recursive by default. So similarly to how you would make a let binding recursive by using the rec keyword:
let rec f = ...
you can make a type definition non-recursive by using the nonrec keyword:
type nonrec t = t
Does "Value Restriction" practically mean that there is no higher order functional programming?
I have a problem that each time I try to do a bit of HOP I get caught by a VR error. Example:
let simple (s:string)= fun rq->1
let oops= simple ""
type 'a SimpleType= F of (int ->'a-> 'a)
let get a = F(fun req -> id)
let oops2= get ""
and I would like to know whether it is a problem of a prticular implementation of VR or it is a general problem that has no solution in a mutable type-infered language that doesn't include mutation in the type system.
Does “Value Restriction” mean that there is no higher order functional programming?
Absolutely not! The value restriction barely interferes with higher-order functional programming at all. What it does do is restrict some applications of polymorphic functions—not higher-order functions—at top level.
Let's look at your example.
Your problem is that oops and oops2 are both the identity function and have type forall 'a . 'a -> 'a. In other words each is a polymorphic value. But the right-hand side is not a so-called "syntactic value"; it is a function application. (A function application is not allowed to return a polymorphic value because if it were, you could construct a hacky function using mutable references and lists that would subvert the type system; that is, you could write a terminating function type type forall 'a 'b . 'a -> 'b.
Luckily in almost all practical cases, the polymorphic value in question is a function, and you can define it by eta-expanding:
let oops x = simple "" x
This idiom looks like it has some run-time cost, but depending on the inliner and optimizer, that can be got rid of by the compiler—it's just the poor typechecker that is having trouble.
The oops2 example is more troublesome because you have to pack and unpack the value constructor:
let oops2 = F(fun x -> let F f = get "" in f x)
This is quite a but more tedious, but the anonymous function fun x -> ... is a syntactic value, and F is a datatype constructor, and a constructor applied to a syntactic value is also a syntactic value, and Bob's your uncle. The packing and unpacking of F is all going to be compiled into the identity function, so oops2 is going to compile into exactly the same machine code as oops.
Things are even nastier when you want a run-time computation to return a polymorphic value like None or []. As hinted at by Nathan Sanders, you can run afoul of the value restriction with an expression as simple as rev []:
Standard ML of New Jersey v110.67 [built: Sun Oct 19 17:18:14 2008]
- val l = rev [];
stdIn:1.5-1.15 Warning: type vars not generalized because of
value restriction are instantiated to dummy types (X1,X2,...)
val l = [] : ?.X1 list
-
Nothing higher-order there! And yet the value restriction applies.
In practice the value restriction presents no barrier to the definition and use of higher-order functions; you just eta-expand.
I didn't know the details of the value restriction, so I searched and found this article. Here is the relevant part:
Obviously, we aren't going to write the expression rev [] in a program, so it doesn't particularly matter that it isn't polymorphic. But what if we create a function using a function call? With curried functions, we do this all the time:
- val revlists = map rev;
Here revlists should be polymorphic, but the value restriction messes us up:
- val revlists = map rev;
stdIn:32.1-32.23 Warning: type vars not generalized because of
value restriction are instantiated to dummy types (X1,X2,...)
val revlists = fn : ?.X1 list list -> ?.X1 list list
Fortunately, there is a simple trick that we can use to make revlists polymorphic. We can replace the definition of revlists with
- val revlists = (fn xs => map rev xs);
val revlists = fn : 'a list list -> 'a list list
and now everything works just fine, since (fn xs => map rev xs) is a syntactic value.
(Equivalently, we could have used the more common fun syntax:
- fun revlists xs = map rev xs;
val revlists = fn : 'a list list -> 'a list list
with the same result.) In the literature, the trick of replacing a function-valued expression e with (fn x => e x) is known as eta expansion. It has been found empirically that eta expansion usually suffices for dealing with the value restriction.
To summarise, it doesn't look like higher-order programming is restricted so much as point-free programming. This might explain some of the trouble I have when translating Haskell code to F#.
Edit: Specifically, here's how to fix your first example:
let simple (s:string)= fun rq->1
let oops= (fun x -> simple "" x) (* eta-expand oops *)
type 'a SimpleType= F of (int ->'a-> 'a)
let get a = F(fun req -> id)
let oops2= get ""
I haven't figured out the second one yet because the type constructor is getting in the way.
Here is the answer to this question in the context of F#.
To summarize, in F# passing a type argument to a generic (=polymorphic) function is a run-time operation, so it is actually type-safe to generalize (as in, you will not crash at runtime). The behaviour of thusly generalized value can be surprising though.
For this particular example in F#, one can recover generalization with a type annotation and an explicit type parameter:
type 'a SimpleType= F of (int ->'a-> 'a)
let get a = F(fun req -> id)
let oops2<'T> : 'T SimpleType = get ""