Is it possible to write "isa" or "match" function on Haskell without TH? It must get constructor and values and to match value with this constructor, but constructor can have multiple arguments. Example:
data X = X a b|Y {age::Int, name::String}
isa c v = ???
goal_of_exercise x y|iff all (isa X) x y = 1
|iff any (isa Y) x y = 2
where iff agg c a b = agg c [a, b]
Or may be exists another solution of goal-of-exercise... ?
The function you're asking for is not just impossible to write - it's absolutely nonsensical in the context of Haskell.
In Haskell all (most of) your data types are indeed, as you've observed, tagged unions carrying type information in runtime. This allows pattern matching to work:
data D = A | B
whichD :: D -> String
whichD A = "it's an A"
whichD B = "it's a B"
This works for constructors of one particular type, though. The signature of whichD specifically says it takes a D, which can be constructed using one of its constructors. If you wanted to extend it to, say, take an analogous D', you could use a sum type for that:
data OneOf = OneD D | OneD' D'
Now, there are cases where this is not the case, namely parametrized contexts:
specialShow :: Show a => a -> String
specialShow a = "it's a special " ++ show a
So now, is it possible to change the behavior of specialShow for some particular as? Yes in principle, but not under that signature. Typically you'd introduce your own type class and provide specific instances for the types you want, and use generic code for all other types you want to bring in. This can pose some problems with overlapping instances, but that's solvable.
What if you insisted on doing the runtime check? I don't think that would even work in all cases, but possibly using some compiler intrinsics you'd be able to force it to tell you more about the actual value. I am pretty sure there's no standard Haskell way to do that, as it's essentially breaking the type system abstraction.
Related
This might seem like a stupid question, but let's say I have a Haskell type class C (a :: K) defined for some fixed kind K, and I know I have instances C a for all types a :: K. Let's furthermore assume I have T :: K -> * and a value t :: C A => T A, is there any way to eliminate the constraint? I feel like this should be possible, but I can't come up with anything.
I'm mainly interested in the case where K is a data kind.
No.
The problem here is simple: foralls in Haskell are parametric polymorphism. This means that the same code must run no matter which type gets chosen. The thing I want to say next needs a way to pull out the C a => part of a type and talk about it in isolation, and there isn't really a way to do that in Haskell's syntax. But making up a type that's kinda like that is pretty standard:
data Dict c where Dict :: c => Dict c
Now the thing you are asking for, essentially, is:
magic :: forall a. Dict (C a)
Unfortunately, the code that you want this to represent -- the behaviors of the class methods -- is different for each type of the appropriate kind.
"Okay, okay," I hear you cry, "but when I write t :: C a => T a, isn't that running different code depending on the choice of a?". But no, it turns out, you aren't. There are two extant implementation schemes for typeclasses: dictionaries (alluded to above, and essentially the same as vtables if you're familiar with that from C++) and explicit type representations. In both cases, under the hood, <C> => <T> is actually a function -- taking the methods implementing the constraint <C> in the one case and a runtime representation of all the types mentioned in the constraint <C> in the other. And that function is the same function in all cases -- the dispatch to different code paths happens in the caller.
The type forall a. Dict (C a), on the other hand, has no such luxury: its type is not permitted to be a function under the hood, but must instead be a value of a specific data type.
Consider this module:
data K = A | B
class C (t::K) where
str :: String
instance C 'A where
str = "A"
instance C 'B where
str = "B"
type family F :: K
foo :: String
foo = str #F
Arguably, C t holds for all t::K, since we have the two instances. Then, however, we define a type family F::K which is, essentially, an unknown kind: it could be A, it could be B. Without a line such type instance F = ... defining F, it is impossible to define foo, since it could be "A" or "B".
Indeed, it is impossible to resolve this constraint.
Note that the type instance might even be defined in another module. Indeed, it is pretty common to define type families and type instances in distinct modules, like we do for type classes and their instances. It is impossible, however, to perform separate compilation when we have no information about the type instance.
The solution is, of course, deferring the constraint resolution by adding the constraint to foo:
foo :: C F => String
foo = str #F
I am a mathematician who works a lot with category theory, and I've been using Haskell for a while to perform certain computations etc., but I am definitely not a programmer. I really love Haskell and want to become much more fluent in it, and the type system is something that I find especially great to have in place when writing programs.
However, I've recently been trying to implement category theoretic things, and am running into problems concerning the fact that you seemingly can't have class method laws in Haskell. In case my terminology here is wrong, what I mean is that I can write
class Monoid c where
id :: c -> c
m :: c -> c -> c
but I can't write some law along the lines of
m (m x y) z == m x $ m y z
From what I gather, this is due to the lack of dependent types in Haskell, but I'm not sure how exactly this is the case (having now read a bit about dependent types). It also seems that the convention is just to include laws like this in comments and hope that you don't accidentally cook up some instance that doesn't satisfy them.
How should I change my approach to Haskell to deal with this problem? Is there a nice mathematical/type-theoretic solution (for example, require the existence of an associator that is an isomorphism (though then the question is, how do we encode isomorphisms without a law?)); is there some 'hack' (using extensions such as DataKinds); should I be drastic and switch to using something like Idris instead; or is the best response to just change the way I think about using Haskell (i.e. accept that these laws can't be implemented in a Haskelly way)?
(bonus) How exactly does the lack of laws come from not supporting dependent types?
You want to require that:
m (m x y) z = m x (m y z) -- (1)
But to require this you need a way to check it. So you, or your compiler (or proof assistant), need to construct a proof of this. And the question is, what type is a proof of (1)?
One could imagine some Proof type but then maybe you could just construct a proof that 0 = 0 instead of a proof of (1) and both would have type Proof. So you’d need a more general type. I can’t decide how to break up the rest of the question so I’ll go for a super brief explanation of the Curry-Howard isomorphism followed by an explanation of how to prove two things are equal and then how dependent types are relevant.
The Curry-Howard isomorphism says that propositions are isomorphic to types and proofs are isomorphic to programs: a type corresponds to a proposition and a proof of that proposition corresponds to a program constructing a value inhabiting that type. Ignoring how many propositions might be expressed as types, an example would be that the type A * B (written (A, B) in Haskell) corresponds to the proposition “A and B,” while the type A + B (written Either A B in Haskell) corresponds to the proposition “A or B.” Finally the type A -> B corresponds to “A implies B,” as a proof of this is a program which takes evidence of A and gives you evidence of B. One should note that there isn’t a way to express not A but one could imagine adding a type Not A with builtins of type Either a (Not a) for the law of the excluded middle as well as Not (Not a) -> a, and a * Not a -> Void (where Void is a type which cannot be inhabited and therefore corresponds to false), but then one can’t really run these programs to get constructivist proofs.
Now we will ignore some realities of Haskell and imagine that there aren’t ways round these rules (in particular undefined :: a says everything is true, and unsafeCoerce :: a -> b says that anything implies anything else, or just other functions that don’t return where their existence does not imply the corresponding proof).
So we know how to combine propositions but what might a proposition be? Well one could be to say that two types are equal. In Haskell this corresponds to the GADT
data Eq a b where Refl :: Eq c c
Where this constructor corresponds to the reflexive property of equality.
[side note: if you’re still interested so far, you may be interested to look up Voevodsky’s univalent foundations, depending on how much the idea of “Homotopy type theory” interests you]
So can we prove something now? How about the transitive property of equality:
trans :: Eq a b -> Eq b c -> Eq a c
trans x y =
case x of
Refl -> -- by this match being successful, the compiler now knows that a = b
case y of
Refl -> -- and now b = c and so the compiler knows a = c
Refl -- the compiler knows that this is of type Eq d d, and as it knows a = c, this typechecks as Eq a c
This feels like one hasn’t really proven anything (especially as this mainly relies on the compiler knowing the transitive and symmetric properties), but one gets a similar feeling when proving simple things in logic as well.
So now how might you prove the original proposition (1)? Well let’s imagine we want a type c to be a monoid then we should also prove that $\forall x,y,z:c, m (m x y) z = m x (m y z).$ So we need a way to express m (m x y) z as a type. Strictly speaking this isn’t dependent types (this can be done with DataKinds to promote values and type families instead of functions). But you do need dependent types to have types depend on values. Specifically if you have a type Nat of natural numbers and a type family Vec :: Nat -> * (* is the kind (read type) of all types) of fixed length vectors, you could define a dependently typed function mkVec :: (n::Nat) -> Vec n. Observe how the type of the output depends on the value of the input.
So your law needs to have functions promoted to type level (skipping the questions about how one defines type equality and value equality), as well as dependent types (made up syntax):
class Monoid c where
e :: c
(*) :: c -> c -> c
idl :: (x::c) -> Eq x (e * x)
idr :: (x::c) -> Eq x (x * e)
assoc :: (x::c) -> (y::c) -> (z::c) -> Eq ((x * y) * z) (x * (y * z))
Observe how types tend to become large with dependent types and proofs. In a language missing typeclasses one could put such values into a record.
Final note on the theory of dependent types and how these correspond to the curry Howard isomorphism.
Dependent types can be considered an answer to the question: what types correspond to the propositions $\forall x\in S\quad P(x)$ and $\exists y\in T\quad Q(y)?$
The answer is that you create new ways to make types: the dependent product and the dependent sum (coproduct). The dependent product expresses “for all values $x$ of type $S,$ there is a value of type $P(x).$” A normal product would be a dependent product with $S=2,$ a type inhabited by two values. A dependent product might be written (x:T) -> P x. A dependent sum says “some value $y$ of type $T$, paired with a value of type $Q(y).$” this might be written (y:T) * Q y.
One can think of these as a generalisation of arbitrarily indexed (co)products from Set to general categories, where one might sensibly write e.g. $\prod_\Lambda X(\lambda),$ and sometimes such notation is used in type theory.
This question already has answers here:
What's so special about 'return' keyword
(3 answers)
Closed 5 years ago.
Consider these functions
f1 :: Maybe Int
f1 = return 1
f2 :: [Int]
f2 = return 1
Both have the same statement return 1. But the results are different. f1 gives value Just 1 and f2 gives value [1]
Looks like Haskell invokes two different versions of return based on return type. I like to know more about this kind of function invocation. Is there a name for this feature in programming languages?
This is a long meandering answer!
As you've probably seen from the comments and Thomas's excellent (but very technical) answer You've asked a very hard question. Well done!
Rather than try to explain the technical answer I've tried to give you a broad overview of what Haskell does behind the scenes without diving into technical detail. Hopefully it will help you to get a big picture view of what's going on.
return is an example of type inference.
Most modern languages have some notion of polymorphism. For example var x = 1 + 1 will set x equal to 2. In a statically typed language 2 will usually be an int. If you say var y = 1.0 + 1.0 then y will be a float. The operator + (which is just a function with a special syntax)
Most imperative languages, especially object oriented languages, can only do type inference one way. Every variable has a fixed type. When you call a function it looks at the types of the argument and chooses a version of that function that fits the types (or complains if it can't find one).
When you assign the result of a function to a variable the variable already has a type and if it doesn't agree with the type of the return value you get an error.
So in an imperative language the "flow" of type deduction follows time in your program Deduce the type of a variable, do something with it and deduce the type of the result. In a dynamically typed language (such as Python or javascript) the type of a variable is not assigned until the value of the variable is computed (which is why there don't seem to be types). In a statically typed language the types are worked out ahead of time (by the compiler) but the logic is the same. The compiler works out what the types of variables are going to be, but it does so by following the logic of the program in the same way as the program runs.
In Haskell the type inference also follows the logic of the program. Being Haskell it does so in a very mathematically pure way (called System F). The language of types (that is the rules by which types are deduced) are similar to Haskell itself.
Now remember Haskell is a lazy language. It doesn't work out the value of anything until it needs it. That's why it makes sense in Haskell to have infinite data structures. It never occurs to Haskell that a data structure is infinite because it doesn't bother to work it out until it needs to.
Now all that lazy magic happens at the type level too. In the same way that Haskell doesn't work out what the value of an expression is until it really needs to, Haskell doesn't work out what the type of an expression is until it really needs to.
Consider this function
func (x : y : rest) = (x,y) : func rest
func _ = []
If you ask Haskell for the type of this function it has a look at the definition, sees [] and : and deduces that it's working with lists. But it never needs to look at the types of x and y, it just knows that they have to be the same because they end up in the same list. So it deduces the type of the function as [a] -> [a] where a is a type that it hasn't bothered to work out yet.
So far no magic. But it's useful to notice the difference between this idea and how it would be done in an OO language. Haskell doesn't convert the arguments to Object, do it's thing and then convert back. Haskell just hasn't been asked explicitly what the type of the list is. So it doesn't care.
Now try typing the following into ghci
maxBound - length ""
maxBound : "Hello"
Now what just happened !? minBound bust be a Char because I put it on the front of a string and it must be an integer because I added it to 0 and got a number. Plus the two values are clearly very different.
So what is the type of minBound? Let's ask ghci!
:type minBound
minBound :: Bounded a => a
AAargh! what does that mean? Basically it means that it hasn't bothered to work out exactly what a is, but is has to be Bounded if you type :info Bounded you get three useful lines
class Bounded a where
minBound :: a
maxBound :: a
and a lot of less useful lines
So if a is Bounded there are values minBound and maxBound of type a.
In fact under the hood Bounded is just a value, it's "type" is a record with fields minBound and maxBound. Because it's a value Haskell doesn't look at it until it really needs to.
So I appear to have meandered somewhere in the region of the answer to your question. Before we move onto return (which you may have noticed from the comments is a wonderfully complex beast.) let's look at read.
ghci again
read "42" + 7
read "'H'" : "ello"
length (read "[1,2,3]")
and hopefully you won't be too surprised to find that there are definitions
read :: Read a => String -> a
class Read where
read :: String -> a
so Read a is just a record containing a single value which is a function String -> a. Its very tempting to assume that there is one read function which looks at a string, works out what type is contained in the string and returns that type. But it does the opposite. It completely ignores the string until it's needed. When the value is needed, Haskell first works out what type it's expecting, once it's done that it goes and gets the appropriate version of the read function and combines it with the string.
now consider something slightly more complex
readList :: Read a => [String] -> a
readList strs = map read strs
under the hood readList actually takes two arguments
readList' (Read a) -> [String] -> [a]
readList' {read = f} strs = map f strs
Again as Haskell is lazy it only bothers looking at the arguments when it's needs to find out the return value, at that point it knows what a is, so the compiler can go and fine the right version of Read. Until then it doesn't care.
Hopefully that's given you a bit of an idea of what's happening and why Haskell can "overload" on the return type. But it's important to remember it's not overloading in the conventional sense. Every function has only one definition. It's just that one of the arguments is a bag of functions. read_str doesn't ever know what types it's dealing with. It just knows it gets a function String -> a and some Strings, to do the application it just passes the arguments to map. map in turn doesn't even know it gets strings. When you get deeper into Haskell it becomes very important that functions don't know very much about the types they're dealing with.
Now let's look at return.
Remember how I said that the type system in Haskell was very similar to Haskell itself. Remember that in Haskell functions are just ordinary values.
Does this mean I can have a type that takes a type as an argument and returns another type? Of course it does!
You've seen some type functions Maybe takes a type a and returns another type which can either be Just a or Nothing. [] takes a type a and returns a list of as. Type functions in Haskell are usually containers. For example I could define a type function BinaryTree which stores a load of a's in a tree like structure. There are of course lots of much stranger ones.
So, if these type functions are similar to ordinary types I can have a typeclass that contains type functions. One such typeclass is Monad
class Monad m where
return a -> m a
(>>=) m a (a -> m b) -> m b
so here m is some type function. If I want to define Monad for m I need to define return and the scary looking operator below it (which is called bind)
As others have pointed out the return is a really misleading name for a fairly boring function. The team that designed Haskell have since realised their mistake and they're genuinely sorry about it. return is just an ordinary function that takes an argument and returns a Monad with that type in it. (You never asked what a Monad actually is so I'm not going to tell you)
Let's define Monad for m = Maybe!
First I need to define return. What should return x be? Remember I'm only allowed to define the function once, so I can't look at x because I don't know what type it is. I could always return Nothing, but that seems a waste of a perfectly good function. Let's define return x = Just x because that's literally the only other thing I can do.
What about the scary bind thing? what can we say about x >>= f? well x is a Maybe a of some unknown type a and f is a function that takes an a and returns a Maybe b. Somehow I need to combine these to get a Maybe b`
So I need to define Nothing >== f. I can't call f because it needs an argument of type a and I don't have a value of type a I don't even know what 'a' is. I've only got one choice which is to define
Nothing >== f = Nothing
What about Just x >>= f? Well I know x is of type a and f takes a as an argument, so I can set y = f a and deduce that y is of type b. Now I need to make a Maybe b and I've got a b so ...
Just x >>= f = Just (f x)
So I've got a Monad! what if m is List? well I can follow a similar sort of logic and define
return x = [x]
[] >>= f = []
(x : xs) >>= a = f x ++ (xs >>= f)
Hooray another Monad! It's a nice exercise to go through the steps and convince yourself that there's no other sensible way of defining this.
So what happens when I call return 1?
Nothing!
Haskell's Lazy remember. The thunk return 1 (technical term) just sits there until someone needs the value. As soon as Haskell needs the value it know what type the value should be. In particular it can deduce that m is List. Now that it knows that Haskell can find the instance of Monad for List. As soon as it does that it has access to the correct version of return.
So finally Haskell is ready To call return, which in this case returns [1]!
The return function is from the Monad class:
class Applicative m => Monad (m :: * -> *) where
...
return :: a -> m a
So return takes any value of type a and results in a value of type m a. The monad, m, as you've observed is polymorphic using the Haskell type class Monad for ad hoc polymorphism.
At this point you probably realize return is not an good, intuitive, name. It's not even a built in function or a statement like in many other languages. In fact a better-named and identically-operating function exists - pure. In almost all cases return = pure.
That is, the function return is the same as the function pure (from the Applicative class) - I often think to myself "this monadic value is purely the underlying a" and I try to use pure instead of return if there isn't already a convention in the codebase.
You can use return (or pure) for any type that is a class of Monad. This includes the Maybe monad to get a value of type Maybe a:
instance Monad Maybe where
...
return = pure -- which is from Applicative
...
instance Applicative Maybe where
pure = Just
Or for the list monad to get a value of [a]:
instance Applicative [] where
{-# INLINE pure #-}
pure x = [x]
Or, as a more complex example, Aeson's parse monad to get a value of type Parser a:
instance Applicative Parser where
pure a = Parser $ \_path _kf ks -> ks a
Imagine a language which doesn't allow multiple value constructors for a data type. Instead of writing
data Color = White | Black | Blue
we would have
data White = White
data Black = Black
data Blue = Black
type Color = White :|: Black :|: Blue
where :|: (here it's not | to avoid confusion with sum types) is a built-in type union operator. Pattern matching would work in the same way
show :: Color -> String
show White = "white"
show Black = "black"
show Blue = "blue"
As you can see, in contrast to coproducts it results in a flat structure so you don't have to deal with injections. And, unlike sum types, it allows to randomly combine types resulting in greater flexibility and granularity:
type ColorsStartingWithB = Black :|: Blue
I believe it wouldn't be a problem to construct recursive data types as well
data Nil = Nil
data Cons a = Cons a (List a)
type List a = Cons a :|: Nil
I know union types are present in TypeScript and probably other languages, but why did the Haskell committee chose ADTs over them?
Haskell's sum type is very similar to your :|:.
The difference between the two is that the Haskell sum type | is a tagged union, while your "sum type" :|: is untagged.
Tagged means every instance is unique - you can distunguish Int | Int from Int (actually, this holds for any a):
data EitherIntInt = Left Int | Right Int
In this case: Either Int Int carries more information than Int because there can be a Left and Right Int.
In your :|:, you cannot distinguish those two:
type EitherIntInt = Int :|: Int
How do you know if it was a left or right Int?
See the comments for an extended discussion of the section below.
Tagged unions have another advantage: The compiler can verify whether you as the programmer handled all cases, which is implementation-dependent for general untagged unions. Did you handle all cases in Int :|: Int? Either this is isomorphic to Int by definition or the compiler has to decide which Int (left or right) to choose, which is impossible if they are indistinguishable.
Consider another example:
type (Integral a, Num b) => IntegralOrNum a b = a :|: b -- untagged
data (Integral a, Num b) => IntegralOrNum a b = Either a b -- tagged
What is 5 :: IntegralOrNum Int Double in the untagged union? It is both an instance of Integral and Num, so we can't decide for sure and have to rely on implementation details. On the other hand, the tagged union knows exactly what 5 should be because it is branded with either Left or Right.
As for naming: The disjoint union in Haskell is a union type. ADTs are only a means of implementing these.
I will try to expand the categorical argument mentioned by #BenjaminHodgson.
Haskell can be seen as the category Hask, in which objects are types and morphisms are functions between types (disregarding bottom).
We can define a product in Hask as tuple - categorically speaking it meets the definition of the product:
A product of a and b is the type c equipped with projections p and q such that p :: c -> a and q :: c -> b and for any other candidate c' equipped with p' and q' there exists a morphism m :: c' -> c such that we can write p' as p . m and q' as q . m.
Read up on this in Bartosz' Category Theory for Programmers for further information.
Now for every category, there exists the opposite category, which has the same morphism but reverses all the arrows. The coproduct is thus:
The coproduct c of a and b is the type c equipped with injections i :: a -> c and j :: b -> c such that for all other candidates c' with i' and j' there exists a morphism m :: c -> c' such that i' = m . i and j' = m . j.
Let's see how the tagged and untagged union perform given this definition:
The untagged union of a and b is the type a :|: b such that:
i :: a -> a :|: b is defined as i a = a and
j :: b -> a :|: b is defined as j b = b
However, we know that a :|: a is isomorphic to a. Based on that observation we can define a second candidate for the product a :|: a :|: b which is equipped with the exact same morphisms. Therefore, there is no single best candidate, since the morphism m between a :|: a :|: b and a :|: b is id. id is a bijection, which implies that m is invertible and "convert" types either way. A visual representation of that argument. Replace p with i and q with j.
Restricting ourselves Either, as you can verify yourself with:
i = Left and
j = Right
This shows that the categorical complement of the product type is the disjoint union, not the set-based union.
The set union is part of the disjoint union, because we can define it as follows:
data Left a = Left a
data Right b = Right b
type DisjUnion a b = Left a :|: Right b
Because we have shown above that the set union is not a valid candidate for the coproduct of two types, we would lose many "free" properties (which follow from parametricity as leftroundabout mentioned) by not choosing the disjoint union in the category Hask (because there would be no coproduct).
This is an idea I've thought a lot about myself: a language with “first-class type algebra”. Pretty sure we could do about everything this way that we do in Haskell. Certainly if these disjunctions were, like Haskell alternatives, tagged unions; then you could directly rewrite any ADT to use them. In fact GHC can do this for you: if you derive a Generic instance, a variant type will be represented by a :+: construct, which is in essence just Either.
I'm not so sure if untagged unions would also do. As long as you require the types participating in a sum to be discernibly different, the explicit tagging should in principle not be necessary. The language would then need a convenient way to match on types at runtime. Sounds a lot like what dynamic languages do – obviously comes with quite some overhead though.
The biggest problem would be that if the types on both sides of :|: must be unequal then you lose parametricity, which is one of Haskell's nicest traits.
Given that you mention TypeScript, it is instructive to have a look at what its docs have to say about its union types. The example there starts from a function...
function padLeft(value: string, padding: any) { //etc.
... that has a flaw:
The problem with padLeft is that its padding parameter is typed as any. That means that we can call it with an argument that’s neither a number nor a string
One plausible solution is then suggested, and rejected:
In traditional object-oriented code, we might abstract over the two types by creating a hierarchy of types. While this is much more explicit, it’s also a little bit overkill.
Rather, the handbook suggests...
Instead of any, we can use a union type for the padding parameter:
function padLeft(value: string, padding: string | number) { // etc.
Crucially, the concept of union type is then described in this way:
A union type describes a value that can be one of several types.
A string | number value in TypeScript can be either of string type or of number type, as string and number are subtypes of string | number (cf. Alexis King's comment to the question). An Either String Int value in Haskell, however, is neither of String type nor of Int type -- its only, monomorphic, type is Either String Int. Further implications of that difference show up in the remainder of the discussion:
If we have a value that has a union type, we can only access members that are common to all types in the union.
In a roughly analogous Haskell scenario, if we have, say, an Either Double Int, we cannot apply (2*) directly on it, even though both Double and Int have instances of Num. Rather, something like bimap is necessary.
What happens when we need to know specifically whether we have a Fish? [...] we’ll need to use a type assertion:
let pet = getSmallPet();
if ((<Fish>pet).swim) {
(<Fish>pet).swim();
}
else {
(<Bird>pet).fly();
}
This sort of downcasting/runtime type checking is at odds with how the Haskell type system ordinarily works, even though it can be implemented using the very same type system (also cf. leftaroundabout's answer). In contrast, there is nothing to figure out at runtime about the type of an Either Fish Bird: the case analysis happens at value level, and there is no need to deal with anything failing and producing Nothing (or worse, null) due to runtime type mismatches.
In Haskell, you can use unsafeCoerce to override the type system. How to do the same in F#?
For example, to implement the Y-combinator.
I'd like to offer a different solution, based on embedding the untyped lambda calculus in a typed functional language. The idea is to create a data type that allows us to change between types α and α → α, which subsequently allows to escape the restrictions of a type system. I'm not very familiar with F# so I'll give my answer in Haskell, but I believe it could be adapted easily (perhaps the only complication could be F#'s strictness).
-- | Roughly represents morphism between #a# and #a -> a#.
-- Therefore we can embed a arbitrary closed λ-term into #Any a#. Any time we
-- need to create a λ-abstraction, we just nest into one #Any# constructor.
--
-- The type parameter allows us to embed ordinary values into the type and
-- retrieve results of computations.
data Any a = Any (Any a -> a)
Note that the type parameter isn't significant for combining terms. It just allows us to embed values into our representation and extract them later. All terms of a particular type Any a can be combined freely without restrictions.
-- | Embed a value into a λ-term. If viewed as a function, it ignores its
-- input and produces the value.
embed :: a -> Any a
embed = Any . const
-- | Extract a value from a λ-term, assuming it's a valid value (otherwise it'd
-- loop forever).
extract :: Any a -> a
extract x#(Any x') = x' x
With this data type we can use it to represent arbitrary untyped lambda terms. If we want to interpret a value of Any a as a function, we just unwrap its constructor.
First let's define function application:
-- | Applies a term to another term.
($$) :: Any a -> Any a -> Any a
(Any x) $$ y = embed $ x y
And λ abstraction:
-- | Represents a lambda abstraction
l :: (Any a -> Any a) -> Any a
l x = Any $ extract . x
Now we have everything we need for creating complex λ terms. Our definitions mimic the classical λ-term syntax, all we do is using l to construct λ abstractions.
Let's define the Y combinator:
-- λf.(λx.f(xx))(λx.f(xx))
y :: Any a
y = l (\f -> let t = l (\x -> f $$ (x $$ x))
in t $$ t)
And we can use it to implement Haskell's classical fix. First we'll need to be able to embed a function of a -> a into Any a:
embed2 :: (a -> a) -> Any a
embed2 f = Any (f . extract)
Now it's straightforward to define
fix :: (a -> a) -> a
fix f = extract (y $$ embed2 f)
and subsequently a recursively defined function:
fact :: Int -> Int
fact = fix f
where
f _ 0 = 1
f r n = n * r (n - 1)
Note that in the above text there is no recursive function. The only recursion is in the Any data type, which allows us to define y (which is also defined non-recursively).
In Haskell, unsafeCoerce has the type a -> b and is generally used to assert to the compiler that the thing being coerced actually has the destination type and it's just that the type-checker doesn't know it.
Another, less common use, is to reinterpret a pattern of bits as another type. For example an unboxed Double# could be reinterpreted as an unboxed Int64#. You have to be sure about the underlying representations for this to be safe.
In F#, the first application can be achieved with box |> unbox as John Palmer said in a comment on the question. If possible use explicit type arguments to make sure that you don't accidentally have the wrong coercion inferred, e.g. box<'a> |> unbox<'b> where 'a and 'b are type variables or concrete types that are already in scope in your code.
For the second application, look at the BitConverter class for specific conversions of bit-patterns. In theory you could also do something like interfacing with unmanaged code to achieve this, but that seems very heavyweight.
These techniques won't work for implementing the Y combinator because the cast is only valid if the runtime objects actually do have the target type, but with the Y combinator you actually need to call the same function again but with a different type. For this you need the kinds of encoding tricks mentioned in the question John Palmer linked to.