Instantiating a case class from a list of paramete

问题:

Given:

case class Foo(a: Int, b: String, c: Double)

you can say:

val params = Foo(1, "bar", 3.14).productIterator.toList

and get:

params: List[Any] = List(1, bar, 3.14)

Is there a way to "go backwards" and recreate a Foo object directly from this list, i.e.:

Foo.createFromList(params)   // hypothetical

instead of writing:

Foo(params(0).asInstanceOf[Int], params(1).asInstanceOf[String], params(2).asInstanceOf[Double])

EDIT: it seems that it boils down to being able to send the elements of a list as parameters to a function without writing them out explicitly, e.g.:

def bar(a: Int, b: Int, c: Int) = //...
val list = List(1, 2, 3, 4, 5)
bar(list.take(3)) // hypothetical, instead of:
bar(list(0), list(1), list(2))

I would sort of expect to be able to do:

bar(list.take(3): _*)

but that doesn't seem to work.

EDIT: Solution based on extempore's answer, but invoking the constructor directly instead of using the apply method:

case class Foo(a: Int = 0, b: String = "bar", c: Double = 3.14) {
    val cs = this.getClass.getConstructors
    def createFromList(params: List[Any]) =
    cs(0).newInstance(params map { _.asInstanceOf[AnyRef] } : _*).asInstanceOf[Foo]
}

Now you can do:

scala> Foo().createFromList(List(4, "foo", 9.81))
res13: Foo = Foo(4,foo,9.81)

You can also refactor the creation method into a trait:

trait Creatable[T <: Creatable[T]] {
    val cs = this.getClass.getConstructors
    def createFromList(params: List[Any]) =
        cs(0).newInstance(params map { _.asInstanceOf[AnyRef] } : _*).asInstanceOf[T]   
}

case class Bar(a: Int = 0, b: String = "bar", c: Double = 3.14) extends Creatable[Bar]

And do e.g.:

scala> val bar = Bar()
bar: Bar = Bar(0,bar,3.14)

scala> bar == bar.createFromList(bar.productIterator.toList)
res11: Boolean = true

回答1:

scala> case class Foo(a: Int, b: String, c: Double)
defined class Foo

scala> val params = Foo(1, "bar", 3.14).productIterator.toList
params: List[Any] = List(1, bar, 3.14)

scala> Foo.getClass.getMethods.find(x => x.getName == "apply" && x.isBridge).get.invoke(Foo, params map (_.asInstanceOf[AnyRef]): _*).asInstanceOf[Foo]
res0: Foo = Foo(1,bar,3.14)

scala> Foo(1, "bar", 3.14) == res0
res1: Boolean = true

Edit: by the way, the syntax so far only being danced around for supplying the tuple as an argument is:

scala> case class Foo(a: Int, b: String, c: Double)
defined class Foo

scala> Foo.tupled((1, "bar", 3.14))                
res0: Foo = Foo(1,bar,3.14)

回答2:

Well, you can certainly do this with a tuple:

(Foo _).tupled apply (1, bar, 3.14)

But there is no real way to get from a List[S] to (A, B, C) for A, B, C <: S. There may be a way of doing this with HLists of course

回答3:

You could use pattern matching like:

params match {                                   
 case List(x:Int, y:String, d:Double) => Foo(x,y,d)
}