如何在Scala中设置多个or类型边界

正如您所说的,这不太可能,但是您可以使用类型类模式来完成。例如,从 here :

sealed abstract class Acceptable[T]
object Acceptable {
  implicit object IntOk extends Acceptable[Int]
  implicit object LongOk extends Acceptable[Long]
}

def f[T: Acceptable](t: T) = t

scala> f(1)
res0: Int = 1

scala> f(1L)
res1: Long = 1

scala> f(1.0)
<console>:8: error: could not find implicit value for parameter ev: Acceptable[Double]
f(1.0)
^

编辑

如果类和对象是伙伴,则此方法有效。在REPL上,如果您在不同的行中键入每一个(即,它们之间出现一个“result”),则它们不是同伴。不过,您可以按以下方式键入:

scala> sealed abstract class Acceptable[T]; object Acceptable {
     |   implicit object IntOk extends Acceptable[Int]
     |   implicit object LongOk extends Acceptable[Long]
     | }
defined class Acceptable
defined module Acceptable

你可以从这两种车中获得一点里程。但是,这两种层次结构都是密封的,处理两种以上的类型会变得很麻烦。

scala> implicit def string2either(s: String) = Left(s)
string2either: (s: String)Left[String,Nothing]

scala> implicit def int2either(i: Int) = Right(i)
int2either: (i: Int)Right[Nothing,Int]

scala> type SorI = Either[String, Int]
defined type alias SorI

scala> def foo(a: SorI) {a match {
     |     case Left(v)  => println("Got a "+v)
     |     case Right(v) => println("Got a "+v)
     |   }
     | }
foo: (a: SorI)Unit

scala> def bar(a: List[SorI]) {
     |   a foreach foo
     | }
bar: (a: List[SorI])Unit

scala>

scala> foo("Hello")
Got a Hello

scala> foo(10)
Got a 10

scala> bar(List(99, "beer"))
Got a 99
Got a beer

case class IntList(l:List[Int])
case class StringList(l:List[String])

implicit def li2il(l:List[Int]) = IntList(l)
implicit def ls2sl(l:List[String]) = StringList(l)

def foo(list:IntList) =  { println("Int-List " + list.l)}
def foo(list:StringList) =  { println("String-List " + list.l)}

有一个黑客:

implicit val x: Int = 0
def foo(a: List[Int])(implicit ignore: Int) { }

implicit val y = ""
def foo(a: List[String])(implicit ignore: String) { }

foo(1::2::Nil)
foo("a"::"b"::Nil)

看到了吗 http://michid.wordpress.com/2010/06/14/working-around-type-erasure-ambiguities-scala/

question .