I wish to implement a stack with value restriction.
What I want is that the pop and push are always talking about exactly the same type all the time.
Here is my sig.
module type MyStackSig =
sig
type 'a stack
exception EmptyStack
val create : unit -> 'a stack
val push : 'a stack -> 'a -> unit
val pop : 'a stack -> 'a
val is_empty : 'a stack -> bool
val size : 'a stack -> int
end;;
Is this sig enough for the value restriction?
I mean will push and pop be talking about the same type all the time?
Your signature looks correct. The types tell that push and pop handle the same content type when they take the same stack. The type parameter 'a of stack ensures it.
let st = create () in
push st 1;
print_string (pop st)
is ill typed, since the st cannot have a polymorphic type due to value restriction, however, it is used for more than one type: int stack and string stack: "they do not talk about the same type" for one stack st, whose parameter type is value-restricted.
On the other hand, the following is well-typed:
let st1 = create () in
let st2 = create () in
push st1 1;
print_string (pop st2) (* it should raise EmptyStack, but do not care it here *)
Here, push and pop talk about different types, but of different stacks. So, no problem.
(Relaxed) value restriction is not something you can force. You are forced to live with it. It is a limitation of the type system for typing of side effects in OCaml.
Related
I have been stuck on this question for a while. I've been editing and reviewing and changing the types for a while but I can't get the type checker to accept what I am doing, probably because I don't fully understand the error/where I am going wrong on this. I am working with the type:
type 'a pred = 'a -> bool
I believe this means I can use 'a pred as a shortcut to mean 'a -> bool, so an int going to result in a bool in my case, but I don't fully get how to implement it because I can't find many examples on this online which I have checked for.
My latest version is below, but I am getting a few errors from the checker, including Error: operator and operand do not agree. Would someone be able to explain where my error is, and why?
Edit: I now think there is a mismatch between this function and the rest of the code. The rest of the code requires it to be an 'a, polymorphic, while here I am assuming it is an int. However, I'm not sure how to do this function (check if odd) while keeping it a polymorphic type.
fun isOdd (p : int) : bool =
case p
of 1 => true
| 0 => false
| _ => isOdd (p - 2)
I believe this means I can use 'a pred as a shortcut to mean 'a -> bool
That is correct.
In the case of your isOdd predicate, it is an int pred:
> val isOdd = fn : int -> bool
- isOdd : int pred;
> val it = fn : int -> bool
Perhaps your misconception lies in the fact that in spite of expressing : int pred, the result in the REPL is still described as int -> bool? This is because we have only defined a type alias, and those tend reduce to their non-aliased form in SML.
Or perhaps your misconception lies in the 'a reducing to some concrete value? You can operate with 'a pred by not referring to concrete values of 'a. For example, if you want to filter an 'a list for only values that are true for a given 'a pred, then the standard function List.filter will have the type:
- List.filter : 'a pred -> 'a list -> 'a list;
> val 'a it = fn : ('a -> bool) -> 'a list -> 'a list
I'm not sure how to do this function (check if odd) while keeping it a polymorphic type.
I'm not sure, either.
Oddness is a property of integers, not arbitrary types 'a.
You would need to extend the meaning of "odd" to any type first. Then you would need some kind of overloading, since the oddness of every type presumably isn't determined by the same mechanism. I'm pretty sure this is a side-track caused by one or two confusions.
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.
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
After learning OCaml for like half year, I am still struggling at the functional programming and imperative programming bits.
It is not about using list or array, but about API design.
For example, if I am about to write stack for a user, should I present it in functional or imperative way?
stack should have a function called pop, which means return the last element to user and remove it from the stack. So if I design my stack in functional way, then for pop, I should return a tuple (last_element, new_stack), right? But I think it is ugly.
At the same time, I feel functional way is more natural in Functional Programming.
So, how should I handle this kind of design problem?
Edit
I saw stack's source code and they define the type like this:
type 'a t = { mutable c : 'a list }
Ok, internally the standard lib uses list which is immutable, but the encapsulate it in a mutable record.
I understand this as in this way, for the user, it is always one stack and so no need for a tuple to return to the client.
But still, it is not a functional way, right?
Mutable structures are sometimes more efficient, but they're not persistent, which is useful in various situations (mostly for backtracking a failed computation). If the immutable interface has no or little performance overhead over the mutable interface, you should absolutely prefer the immutable one.
Functionally (i.e. without mutability), you can either define it exactly like List by using head/tail rather than pop, or you can, as you suggest, let the API handle state change by returning a tuple. This is comparable to how state monads are built.
So either it is the responsibility of the parent scope to handle the stack's state (e.g. through recursion), in which case stacks are exactly like lists, or some of this responsibility is loaded to the API through tuples.
Here is a quick attempt (pretending to know O'Caml syntax):
module Stack =
struct
type 'a stack = 'a list
let empty _ = ((), [])
let push x stack = ((), x::stack)
let pop (x::stack) = (x, stack)
| pop _ = raise EmptyStack
end
One use case would then be:
let (_, st) = Stack.empty ()
let (_, st) = Stack.push 1 Stack.empty
let (_, st) = Stack.push 2 st
let (_, st) = Stack.push 3 st
let (x, st) = Stack.pop st
Instead of handling the tuples explicitly, you may want to hide passing on st all the time and invent an operator that makes the following syntax possible:
let (x, st) = (Stack.empty >>= Stack.push 1 >>=
Stack.push 2 >>= Stack.push 3 >>= Stack.pop) []
If you can make this operator, you have re-invented the state monad. :)
(Because all of the functions above take a state as its curried last argument, they can be partially applied. To expand on this, so it is more apparent what is going on, but less readable, see the rewrite below.)
let (x, st) = (fun st -> Stack.empty st >>= fun st -> Stack.push 1 st
>>= fun st -> Stack.push 2 st
>>= fun st -> Stack.push 3 st
>>= fun st -> Stack.pop) []
This is one idiomatic way to deal with state and immutable data structures, at least.
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 ""