按照你的要求说不太可能,但可以使用类型类模式来实现。例如,可以从这里学习:
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
如果类和对象是伴侣关系,则此方法有效。在REPL中,如果您将它们分别输入不同的行(即,它们之间出现一个“结果”),则它们不是伴侣。但您可以像下面这样输入:
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
Either类型可能会有一些用处。然而,Either的层次结构是封闭的,处理超过两种类型就变得麻烦了。
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
另一种解决方案是包装类(wrapper classes):
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)}
有一个黑客技巧:
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)
请查看http://michid.wordpress.com/2010/06/14/working-around-type-erasure-ambiguities-scala/。
还有这个问题。