Why is Scala's behavior in case of overloading

2019-04-22 17:47发布

Given this Scala code:

object test {

  def byval(a: Int) = println("Int")
  def byval(a: Long) = println("Long")

  def byname(a: => Int) = println("=> Int")
  def byname(a: => Long) = println("=> Long")

  def main(args: Array[String]) {
      byval(5)
      byname(5)
  }
}

the call byval(5) compiles correctly, but byname fails to compile:

ambiguous reference to overloaded definition

Why? I would expect to observe the same behavior for by-value and by-name parameters with respect to overloading… How can it be fixed?

2条回答
霸刀☆藐视天下
2楼-- · 2019-04-22 18:31

Possible workaround without overloading (in addition to what has been said earlier), if you don't want to use different method names:

def byname[A](a: => A)(implicit manifest:Manifest[A]) = 
manifest.erasure match {
   case erasure if(erasure.isAssignableFrom(classOf[Long])) => println("=> Long")
   case erasure if(erasure.isAssignableFrom(classOf[Int])) => println("=> Int")
}
查看更多
劫难
3楼-- · 2019-04-22 18:37

That's because JVM does not support a "by-name" parameter, so Scala has to implement it in another way. => X actually compiles to a Function0[X], which erases to Function0[Object], which makes it impossible for Scala to distinguish two methods that differ only by the expected type of a by-name parameter.

查看更多
登录 后发表回答