I read a program with the following definition:
type 'a queue = ('a list * 'a list) ref
I do not understand the syntax here. Is it a union or what?
This is a synonym/alias. The type ('a list * 'a list) ref already makes sense in OCaml (it is a reference to a pair of lists of type 'a), we are giving it a new, shorter name, 'a queue. More precisely, queue is the name of a parametrized type, and the parameter is named 'a here.
Related
when I write a type that e.g. only accepts strings: let type t1 = string
I can do let name : t1 = "A", but not let age : t1 = 1
But when I want to have a generic type that accepts any data type I have to do this: let type 'a t2 = 'a So I can do both let name : t2 = "A", let age : t2 = 1.
But why do I have to write let type 'a t2 = 'a instead of let type t2 = 'a?
The form
let type t1 = string
is not syntactically valid OCaml.
I imagine that you meant:
type t1 = string
Similarly, with the type constructor t2 defined as
type 'a t2 = 'a
then
let x : t2 = "hi"
is a type error because t2 is not a type but a type constructor of arity one.
The closest valid definition would be:
let x: string t2 = "hi"
which is equivalent
let x: 'a t2 = "hi"
because the type variable 'a is equated to 'a = string when inferring the type of x. But 'a t2 is an abbreviation for 'a, thus the above is still the same as
let x : string = "hi"
At a higher level, there is no useful generic type that accepts any data¹. Indeed, if it existed such type would break the type system.
¹ There are an advanced feature (GADTs or record with polymorphic fields) that allows to either create black-hole types that can carry a data of any kind but forbids any use of the data or types without any values of this type. However, it is probably better to first familiarize yourself with the core part of the type system before exploring those area.
To add to #octachron's answer, you can think of a declaration like type 'a t = 'a as the declaration of a function over types: given a type 'a, t will produce a new type 'a t (which in this case is just 'a). So you need the 'a just like you would need to declare the parameters in a function.
This shouldn't be confused with the occurrences of 'a in an explicit typing form like expr : 'a t. In this case, OCaml thinks of 'a as a type variable, which may or may not denote polymorphism. If you actually wanted to have polymorphism as is the polymorphic type, you'd have to introduce a polymorphic type variable either with 'a. ... or with a type a:
let id : 'a. 'a -> 'a = fun x -> x
let id (type a) (x : a) : a = x
However, note that these forms are not quite equivalent, there are subtle differences, and they are in any case different (in general, not in this specific example) from let id : 'a -> 'a = fun x -> x or let id (x : 'a) : 'a = x.
The following type declarations do not work:
type 'a or_null = [ 'a | `Null ]
and
type 'a or_null = [ 'a | `Null ] constraint 'a = [> `A | `B ]
With the message:
Error: The type 'a does not expand to a polymorphic variant type
Hint: Did you mean `a
I would like to achieve this without using another layer in the memory representation (and in the syntax). In particular, I want to avoid using an option type such as
type 'a or_null = | A of 'a | Null
Is there a way to have such a type using only polymorphic variants? The final goal would be to write e.g. monads on 'a or_null types. (And this is actually the tricky part.)
Polymorphic variants cannot track the absence of a specific constructor. This implies that we cannot really write the usual bind. If we try
let bind x f =
match x with
| `Null -> `Null
| x -> f x
we get
val bind: ([> `Null] as 'a) -> ('a -> ([>`Null] as 'b)) -> 'b
If for readability's sake, we add the following type abbreviation
type 'a m = [> `Null] as 'a
(which is an alternative definition of or_null) the previous type read as
val bind: 'a m -> ('a m -> 'b m) -> 'b m
In other words, the function argument f of bind must already handle the `Null case in its argument by itself because the type system cannot express the constraint x <> `Null in the second branch of the match.
If I define
fun id x = x
Then naturally id has type 'a -> 'a
Of course, id 0 evaluates to 0, which makes perfect sense.
Since this makes perfect sense, I should be able to encapsulate it by a function:
fun applyToZero (f: 'a -> 'a) = f 0
With the hope that applyToZero will have type ('a -> 'a) -> int and applyToZero id will evaluate to 0
But when I try to define applyToZero as above, SML/NJ gives an odd error message which begins:
unexpected exception (bug?) in SML/NJ: Match [nonexhaustive match failure]
raised at: ../compiler/Elaborator/types/unify.sml:84.37
This almost looks like a bug in the compiler itself. Weird, but possible.
But PolyML doesn't like it either (though its error message is less odd):
> fun applyToZero (f: 'a -> 'a) = f 0;
poly: : error: Type error in function application.
Function: f : 'a -> 'a
Argument: 0 : int
Reason: Can't unify int to 'a (Cannot unify with explicit type variable)
Found near f 0
The following does work:
fun ignoreF (f: 'a -> 'a) = 1
with the inferred type ('a -> 'a) -> int. This shows that it isn't impossible to create a higher order function of this type.
Why doesn't SML accept my definition of applyToZero? Is there any workaround that will allow me to define it so that its type is ('a -> 'a) -> int?
Motivation: in my attempt to solve the puzzle in this question, I was able to define a function tofun of type int -> 'a -> 'a and another function fromfun with the desired property that fromfun (tofun n) = n for all integers n. However, the inferred type of my working fromfun is ('int -> 'int) -> 'int). All of my attempts to add type annotations so that SML will accept it as ('a -> 'a) -> int have failed. I don't want to show my definition of fromfun since the person that asked that question might still be working on that puzzle, but the definition of applyToZero triggers exactly the same error messages.
It can't be done in plain Hindley-Milner, like used by SML, because it does not support so-called higher-ranked or first-class polymorphism. The type annotation 'a -> 'a and the type ('a -> 'a) -> int do not mean what you think they do.
That becomes clearer if we make the binder for the type variable explicit.
fun ignoreF (f: 'a -> 'a) = 1
actually means
fun 'a ignoreF (f: 'a -> 'a) = 1
that is, 'a is a parameter to the whole function ignoreF, not to its argument f. Consequently, the type of the function is
ignoreF : ∀ 'a. (('a -> 'a) -> int)
Here, I make the binder for 'a explicit in the type as a universal quantifier. That's how you write such types in type theory, while SML keeps all quantifiers implicit in its syntax. Now the type you thought this had would be written
ignoreF : (∀ 'a. ('a -> 'a)) -> int
Note the difference: in the first version, the caller of ignoreF gets to choose how 'a is instantiated, hence it could be anything, and the function cannot assume its int (which is why applyToZero does not type-check). In the second type, the caller of the argument gets to choose, i.e., ignoreF.
But such a type is not supported by Hindley-Milner. It only supports so-called prenex polymorphism (or rank 0 polymorphism) where all the ∀ are on the outermost level -- which is why it can keep them implicit, since there is no ambiguity under this restriction. The problem with higher-ranked polymorphism is that type inference for it is undecidable.
So your applyToZero cannot have the type you want in SML. The only way to achieve something like it is by using the module system and its functors:
functor ApplyToZero (val f : 'a -> 'a) = struct val it = f 0 end
Btw, the error message you quote from SML/NJ cannot possibly be caused by the code you showed. You must have done something else.
If we use Hindley-Milner type inference algorithm on fun applyToZero f = f 0 we are going to get f : int -> 'a because of the term f 0.
Obviously, f is a function f : 'b -> 'a. We apply this function to 0, thus 'b = int. Hence, the explicit type annotation f : 'a -> 'a produces the error you observe.
By the way, SML/NJ v110.80 works fine on my machine and prints the following error message:
stdIn:2.39-2.42 Error: operator and operand don't agree [overload - user bound tyvar]
operator domain: 'a
operand: [int ty]
in expression:
f 0
Is there a way to declare something like
type do = ('a -> 'b)
in OCaml? Specifically, to declare a function signature as a type
For free types 'a and 'b,'a -> 'b is not the type of any well behaved OCaml function, because it requires the function to produce a value of an arbitrary type.
So, you can't give a name to a type with unbound parameters:
# type uabfun = 'a -> 'b
Error: Unbound type parameter 'a
If you use specific types, there's no problem giving it a name:
# type iifun = int -> int;;
type iifun = int -> int
If type types 'a and 'b are parameters (rather than being free), there is also no problem:
# type ('a, 'b) abfun = 'a -> 'b;;
type ('a, 'b) abfun = 'a -> 'b
The OCaml manual describes the "constraint" keyword, which can be used in a type definition. However, I cannot figure out any usage that can be done with this keyword. When is this keyword is useful? Can it be used to remove polymorphic type variables? (so that a type 'a t in a module becomes just t and the module can be used in a functor argument which requires t with no variables.)
So, the constraint keywords, used in type or class definitions, let one "reduce the scope” of applicable types to a type parameter, so to speak. The documentation clearly announce that type expressions from both sides of the constraint equation will be unified to "refine" the types the constraint relates to. Because they are type expressions, you may use all the usual type level operators.
Examples:
# type 'a t = int * 'a constraint 'a * int = float * int;;
type 'a t = int * 'a constraint 'a = float
# type ('a,'b) t = 'c r constraint 'c = 'a * 'b
and 'a r = {v1 : 'a; v2 : int };;
type ('a,'b) t = ('a * 'b) r
and 'a r = { v1 : 'a; v2 : int; }
Observe how type unification simplifies the equations, in the first example by getting rid of the extraneous type product (* int), and in the second case eliminating it altogether. Note also that I used a type variable 'c which only appears in the right hand side of the type definition.
Two interesting uses are with polymorphic variants and class types, both based on row-polymorphism. Constraints allow to express certain subtyping relations. By subtyping, for variants, we mean a relation such that any constructor of a type is present in its subtypes. Some of these relations may already be expressed monomorphically:
# type sum_op = [ `add | `subtract ];;
type sum_op = [ `add | `subtract ]
# type prod_op = [ `mul | `div ];;
type prod_op = [ `mul | `div ]
# type op = [ sum_op | prod_op ];;
type op = [ `add | `div | `mul | `sub ]
There, op is a subtype of both sum_op and prod_op.
But in some cases, you have to introduce polymorphism, and this is where constraints come handy:
# type 'a t = 'a constraint [> op ] = 'a;;
type 'a t = 'a constraint 'a = [> op ]
The above let you denote the family of types which are subtypes of op : the type instance is 'a itself for a given instance of 'a t.
If we try to define the same type without a parameter, the type unification algorithm will complain:
# type t' = [> op];;
Error: A type variable is unbound in this type declaration.
In type [> op ] as 'a the variable 'a is unbound
The same sort of constraints may be expressed with class types, and the same problem may arise if the type definition is implicitly polymorphic by subtyping.
# class type ct = object method v : int end;;
class type ct = object method v : int end
# type i = #ct;;
Error: A type variable is unbound in this type declaration.
In type #ct as 'a the variable 'a is unbound
# type 'a i = 'a constraint 'a = #ct;;
type 'a i = 'a constraint 'a = #ct