Scala "does not take parameters" when pattern-matching parametric case-class - scala

I have the following code:
sealed trait A
case class B[T](v: T) extends A
case class C[T](v: T) extends A
object Test {
def swap(a: A): A = a match {
case a: B[t] => C[t](a.v) // works!
case C[t](v) => B[t](v) // error: C[t] does not take parameters
}
}
I would expect either both cases to fail or both of them to work. What's the meaning of the error for the second case? Is there a syntax destructuring parametric case-classes?
Note: Here, 't' in lower case is essential. If it were 'T', the checker would be looking for it in the type parameters of the method.

When you do a match { case C(v) => ??? }, you actually call unapply method of the C companion object, something like this: C.unapply(a) match {Some(v) => ???}
There is only one C object, not an entire family of C[t]. There is no object you can refer to as C[Int], so case C[t](v) => doesn't make sense.
In your example, you use B[t] as a type, not as a pattern, and that's why it works. Note that while the match may succeed, you won't get anything in t, because of type erasure.
When you call C[t](a.v), then first of all, compiler erases type t anyway, and second, this is rewritten to a call to apply method on the companion object: C.apply[t](a.v). Note that the type parameter is on the method call, not on the object.

Simply put, it is not part of the language.
When used in this position, the compiler is looking for the type in the environment, as if it were a usual uppercase type. The type capturing that you are trying to do seems to only work in the case x:Y[z] => ... form of a case statement.
Type capturing in this fashion is not a well know part of the language and had me running to the Scala reference document for the details (Section 8.3). I personally find the that the distinction between upper and lowercase here not to my liking.
In your example t takes on the type of Any as the type information for the parameter is not available from A.

Related

Why does a match on covariant enum does not behave the same as with a sealed trait?

Why does this produce an error?
import scala.io.StdIn
enum Flow[+A] {
case Say (text: String) extends Flow[Unit]
case Ask extends Flow[String]
}
def eval [T](flow: Flow[T]): T = flow match {
case Flow.Say(text) => println(text)
case Flow.Ask => StdIn.readLine // error here
}
18 | case Flow.Ask => StdIn.readLine
| ^^^^^^^^^^^^^^
| Found: String
| Required: T
(no error if the enum is left invariant on T)
But when the enum is replaced with a (seemingly equivalent?) ADT implemented with a sealed trait, no such error is reported:
sealed trait Flow[+A]
object Flow {
case class Say(text: String) extends Flow[Unit]
case object Ask extends Flow[String]
}
// match statement in eval works fine
We can look at the initial proposal which added enums to Scala 3 for some insight.
By the wording of this proposal, enum cases fall into three categories.
Class cases are those cases that are parameterized, either with a type parameter section [...] or with one or more (possibly empty) parameter sections (...).
Simple cases are cases of a non-generic enum class that have neither parameters nor an extends clause or body. That is, they consist of a name only.
Value cases are all cases that do not have a parameter section but that do have a (possibly generated) extends clause and/or a body.
Let's take a look at your enum declaration.
enum Flow[+A] {
case Say (text: String) extends Flow[Unit]
case Ask extends Flow[String]
}
Now, simple cases are right out, because your enum is generic. Your Say is pretty clearly a class case, since it's parameterized. Ask, on the other hand, is non-generic and non-enumerated, so it's a value case. If we scroll down a bit, we see what value cases do
(6) A value case
case C extends <parents> <body>
expands to a value definition
val C = new <parents> { <body>; def enumTag = n; $values.register(this) }
where n is the ordinal number of the case in the companion object, starting from 0. The statement $values.register(this) registers the value as one of the enumValues of the enumeration (see below). $values is a compiler-defined private value in the companion object.
The bottom line here is that Ask is not defined as
object Ask extends Flow[String]
but more like
val Ask = new Flow[String]() { ... }
So your pattern match
case Flow.Ask => StdIn.readLine
is actually an equality check against the value Flow.Ask, rather than a type check against a singleton object. The former, unfortunately, proves nothing to the compiler about the generic type of your value, so it remains T, as opposed to specializing to String.
The rationale for this is provided on the same page
Objectives
...
Enumerations should be efficient, even if they define many values. In particular, we should avoid defining a new class for every value.
...
So it seems the Scala developers wanted to avoid the bloat that would result from an enum declaring tons of unparameterized values (i.e. a type that looks like an ordinary Java-style enum).
The simple solution, then, is to provide a parameter list
case Ask() extends Flow[String]
It's ever so slightly more cumbersome to work with, but it does define a new type, and then your pattern match
case Flow.Ask() => StdIn.readLine
will succeed

Difference between type inference of method and class type parameters in pattern matching

Why does pattern matching work differently when type parameter comes from an enclosing method as opposed to an enclosing class? For example,
trait Base[T]
case class Derived(v: Int) extends Base[Int]
class Test[A] {
def method(arg: Base[A]) = {
arg match {
case Derived(_) => 42
}
}
}
gives error
constructor cannot be instantiated to expected type;
found : A$A87.this.Derived
required: A$A87.this.Base[A]
case Derived(_) => 42
^
whilst it successfully compiles when A is method type parameter
class Test {
def method[A](arg: Base[A]) = {
arg match {
case Derived(_) => 42
}
}
}
The question is based on Daniel's analysis, which I used to attempt to provide answer to similar question.
I don't have the 100% complete answer, but I have a pointer that might be sufficient for you.
Scala compiler deals with GADTs (Generalized Algebraic Data Types) in a very particular way. Some cases are solved with special handling, some cases are unsolved. Dotty is trying to fill most of the holes, and it has already solved a lot of related issues, however there are still quite a few open ones as well.
Typical example of special GADT handling in Scala 2 compiler is very related to your use case. If we take a look at:
def method[A](arg: Base[A]) = {
arg match {
case Derived(_) => 42
}
}
and we explicitly declare the return type to be A:
def method[A](arg: Base[A]): A
it will compile just fine. Your IDE might complain, but the compiler will let it through. Method says it returns an A, but the pattern matching case evaluates into an Int, which theoretically shouldn't compile. However, special handling of GADTs in the compiler says it's fine, because in that particular pattern matching branch A has been "fixed" to be an Int (because we matched on Derived which is a Base[Int]).
Generic type parameter for the GADT (in our case A) has to be declared somewhere. And here's the interesting part - special compiler handling only works when it's declared as the type parameter of the enclosing method. If it's coming from a type member or a type parameter of the enclosing trait/class, it doesn't compile, as you witnessed yourself.
This is why I said it's not a 100% complete answer - I cannot point to a concrete place (such as official specification) which documents this properly. Sources on handling of GADTs in Scala come down to a couple of blogposts, which are great by the way, but if you want more than that you will have to dig into the compiler code yourself. I tried doing exactly that, and I think it comes down to this method, but if you really want to go deeper, you might want to ping someone more experienced with the Scala compiler codebase.

Why can't I apply pattern-matching against non-case classes?

Is there a better explanation than "this is how it works". I mean I tried this one:
class TestShortMatch[T <: AnyRef] {
def foo(t: T): Unit = {
val f = (_: Any) match {
case Val(t) => println(t)
case Sup(l) => println(l)
}
}
class Val(t: T)
class Sup(l: Number)
}
and compiler complaints:
Cannot resolve symbol 'Val'
Cannot resolve symbol 'Sup'
Of course if I add case before each of the classes it will work fine. But what is the reason? Does compiler make some optimization and generate a specific byte-code?
The reason is twofold. Pattern matching is just syntactic sugar for using extractors and case classes happen to give you a couple methods for free, one of which is an extractor method that corresponds to the main constructor.
If you want your example above to work, you need to define an unapply method inside objects Val and Sup. To do that you'd need extractor methods (which are only defined on val fields, so you'll have to make your fields vals):
class Val[T](val t: T)
class Sup(val l: Number)
object Val {
def unapply[T](v: Val[T]): Option[T] = Some(v.t)
}
object Sup {
def unapply(s: Sup): Option[Number] = Some(s.l)
}
And which point you can do something like val Val(v) = new Val("hi"). More often than not, though, it is better to make your class a case class. Then, the only times you should be defining extra extractors.
The usual example (to which I can't seem to find a reference) is coordinates:
case class Coordinate(x: Double, val: Double)
And then you can define a custom extractors like
object Polar {
def unapply(c: Coordinate): Option[(Double,Double)] = {...}
}
object Cartesian {
def unapply(c: Coordinate): Option[(Double,Double)] = Some((c.x,c.y))
}
to convert to the two different representations, all when you pattern match.
You can use pattern matching on arbitrary classes, but you need to implement an unapply method, used to "de-construct" the object.
With a case class, the unapply method is automatically generated by the compiler, so you don't need to implement it yourself.
When you write match exp { case Val(pattern) => ... case ... }, that is equivalent to something like this:
match Val.unapply(exp) {
case Some(pattern) =>
...
case _ =>
// code to match the other cases goes here
}
That is, it uses the result of the companion object's unapply method to see whether the match succeeded.
If you define a case class, it automatically defines a companion object with a suitable unapply method. For a normal class it doesn't. The motivation for that is the same as for the other things that gets automatically defined for case classes (like equals and hashCode for example): By declaring a class as a case class, you're making a statement about how you want the class to behave. Given that, there's a good chance that the auto generated will do what you want. For a general class, it's up to you to define these methods like you want them to behave.
Note that parameters for case classes are vals by default, which isn't true for normal classes. So your class class Val(t: T) doesn't even have any way to access t from the outside. So it isn't even possible to define an unapply method that gets at the value of t. That's another reason why you don't get an automatically generated unapply for normal classes: It isn't even possible to generate one unless all parameters are vals.

What does `SomeType[_]` mean in scala?

Going through Play Form source code right now and encountered this
def bindFromRequest()(implicit request: play.api.mvc.Request[_]): Form[T] = {
I am guessing it takes request as an implicit parameter (you don't have to call bindFromRequet(request)) of type play.api.mvc.Request[_] and return a generic T type wrapped in Form class. But what does [_] mean.
The notation Foo[_] is a short hand for an existential type:
Foo[A] forSome {type A}
So it differs from a normal type parameter by being existentially quantified. There has to be some type so your code type checks where as if you would use a type parameter for the method or trait it would have to type check for every type A.
For example this is fine:
val list = List("asd");
def doSomething() {
val l: List[_] = list
}
This would not typecheck:
def doSomething[A]() {
val l: List[A] = list
}
So existential types are useful in situations where you get some parameterized type from somewhere but you do not know and care about the parameter (or only about some bounds of it etc.)
In general, you should avoid existential types though, because they get complicated fast. A lot of instances (especially the uses known from Java (called wildcard types there)) can be avoided be using variance annotations when designing your class hierarchy.
The method doesn't take a play.api.mvc.Request, it takes a play.api.mvc.Request parameterized with another type. You could give the type parameter a name:
def bindFromRequest()(implicit request: play.api.mvc.Request[TypeParameter]): Form[T] = {
But since you're not referring to TypeParameter anywhere else it's conventional to use an underscore instead. I believe the underscore is special cased as a 'black hole' here, rather than being a regular name that you could refer to as _ elsewhere in the type signature.

Scala: please explain the generics involved

Could some one please explain the generics involved in the following code from play framework
class AuthenticatedRequest[A, U](val user: U, request: Request[A]) extends WrappedRequest[A](request)
class AuthenticatedBuilder[U](userinfo: RequestHeader => Option[U],
onUnauthorized: RequestHeader => Result = _ => Unauthorized(views.html.defaultpages.unauthorized()))
extends ActionBuilder[({ type R[A] = AuthenticatedRequest[A, U] })#R]
The ActionBuilder actualy has type R[A], it is getting reassigned, this much I understand. please explain the intricacies of the syntax
The bit that's confusing you is called a "type lambda". If you search for "scala type lambda", you'll find lots of descriptions and explanations. See e.g. here, from which I'm drawing a lot of inspiration. (Thank you Bartosz Witkowski!)
To describe it very simply, you can think of it as a way to provide a default argument to a type constructor. I know, huh?
Let's break that down. If we have...
trait Unwrapper[A,W[_]] {
/* should throw an Exception if we cannot unwrap */
def unwrap( wrapped : W[A] ) : A
}
You could define an OptionUnwrapper easily enough:
class OptionUnwrapper[A] extends Unwrapper[A,Option] {
def unwrap( wrapped : Option[A] ) : A = wrapped.get
}
But what if we want to define an unwrapper for the very similar Either class, which takes two type parameters [A,B]. Either, like Option, is often used as a return value for things that might fail, but where you might want to retain information about the failure. By convention, "success" results in a Right object containing a B, while failure yields a Left object containing an A. Let's make an EitherUnwrapper, so we have an interface in common with Option to unwrap these sorts of failable results. (Potentially even useful!)
class EitherUnwrapper[A,B] extends Unwrapper[B,Either] { // unwrap to a successful result of type B
def unwrap( wrapped : Either[A,B] ) : B = wrapped match {
case Right( b ) => b // we ignore the left case, allowing a MatchError
}
}
This is conceptually fine, but it doesn't compile! Why not? Because the second parameter of Unwrapper was W[_], that is a type that accepts just one parameter. How can we "adapt" Either's type constructor to be a one parameter type? If we needed a version of an ordinary function or constructor with fewer arguments, we might supply default arguments. So that's exactly what we'll do.
class EitherUnwrapper[A,B] extends Unwrapper[B,({type L[C] = Either[A,C]})#L] {
def unwrap( wrapped : Either[A,B] ) : B = wrapped match {
case Right( b ) => b
}
}
The type alias part
type L[C] = Either[A,C]
adapts Either into a type that requires just one type parameter rather than two, by supplying A as a default first type parameter. But unfortunately, scala doesn't allow you to define type aliases just anywhere: they have to live in a class, trait, or object. But if you define the trait in an enclosing scope, you might not have access to the default value you need for type A! So, the trick is to define a throwaway inner class in a place where A is defined, just where you need the new type.
A set of curly braces can (depending on context) be interpreted as a type definition in scala, for a structural type. For instance in...
def destroy( rsrc : { def close() } ) = rsrc.close()
...the curly brace defines a structural type meaning any object with a close() function. Structural types can also include type aliases.
So { type L[C] = Either[A,C] } is just the type of any object that contains the type alias L[C]. To extract an inner type from an enclosing type -- rather than an enclosing instance -- in Scala, we have to use a type projection rather than a dot. The syntax for a type projection is EnclosingType#InnerType. So, we have { type L[C] = Either[A,C] }#L. For reasons that elude me, the Scala compiler gets confused by that, but if we put the type definition in parentheses, everything works, so we have ({ type L[C] = Either[A,C] })#L.
Which is pretty precisely analogous to ({ type R[A] = AuthenticatedRequest[A, U] })#R in your question. ActionBuilder needs to be parameterized with a type that takes one parameter. AuthenticatedRequest takes two parameters. To adapt AuthenticatedRequest into a type suitable for ActionBuilder, U is provided as a default parameter in the type lambda.