Case classes, pattern matching and curried constructors in Scala
They don't seem to mix that well:
abstract class A
case class B (var a: Int)(var b: String) extends A
case class C extends A
The following will not work:
B(1)("1") match {
case B(a)(b) => print("B")
case C() => print("C")
}
The problem is that pattern matching and curried argumen开发者_JAVA技巧ts do not seem to work. Is there a work-around for this?
If you look at the signature of the unapply function created for the class B, you will see that it is: unapply(x$0: Q): Option[Int]
. Thus, the unapply function works with the first range of parameter of the case classes.
It is confirmed by the scala specification (§5.3.2):
The formal parameters in the first parameter section of a case class are called elements; they are treated specially. First, the value of such a parameter can be extracted as a field of a constructor pattern.
It claims clearly tha only the first parameter section is available through the extractor.
Several workarounds:
- uncurry your parameters
- use a pattern matching with guard if you want to test the 2 values:
case x@B(3) if x.b == "bazinga" => ...
- use a normal class and define your own companion object with your own apply / unapply
What's wrong with this?
def m(a: A) = a match {
case b: B => print("B")
case c: C => print("C")
}
I'm only asking because you didn't ask for more functionality than this.
EDIT
This could help:
object Dog {
def apply(name: String)(size: Int) = new Dog(name)(size)
def unapply(dog: Dog) = Some(dog.name, dog.size)
}
class Dog(val name: String)(var size: Int)
Now you can create dogs either like this:
new Dog("Snoopy")(10)
or like this:
Dog("Snoopy")(10)
But when you pattern match on dogs the constructor pattern is not curried.
Dog("Snoopy")(10) match {
case Dog(a, b) => // do sth with a or b
}
You could use a normal case class and just define a factory method with more than one parameter list.
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