为什么Scala同时拥有unapply
和unapplySeq
?它们之间有什么区别?我应该在什么情况下更倾向于使用其中一个?
为什么Scala同时拥有unapply
和unapplySeq
?它们之间有什么区别?我应该在什么情况下更倾向于使用其中一个?
不详细深入地解释,简化一下:
对于常规参数,apply
构造函数和 unapply
解构函数会被使用:
object S {
def apply(a: A):S = ... // makes a S from an A
def unapply(s: S): Option[A] = ... // retrieve the A from the S
}
val s = S(a)
s match { case S(a) => a }
对于重复的参数,apply
构造函数和unapplySeq
函数进行解构:
object M {
def apply(a: A*): M = ......... // makes a M from an As.
def unapplySeq(m: M): Option[Seq[A]] = ... // retrieve the As from the M
}
val m = M(a1, a2, a3)
m match { case M(a1, a2, a3) => ... }
m match { case M(a, as @ _*) => ... }
请注意,对于第二种情况,重复的参数会被视为一个Seq
,且A*
与_*
之间的相似性也是如此。unapply
。如果您想要解构一个包含Seq
的东西,请使用unapplySeq
。定长参数与可变参数。在《Scala 模式匹配》(pdf)中有很好的解释和对应的示例。我也在这个答案中提供了相应的示例。
简言之:
object Sorted {
def unapply(xs: Seq[Int]) =
if (xs == xs.sortWith(_ < _)) Some(xs) else None
}
object SortedSeq {
def unapplySeq(xs: Seq[Int]) =
if (xs == xs.sortWith(_ < _)) Some(xs) else None
}
scala> List(1,2,3,4) match { case Sorted(xs) => xs }
res0: Seq[Int] = List(1, 2, 3, 4)
scala> List(1,2,3,4) match { case SortedSeq(a, b, c, d) => List(a, b, c, d) }
res1: List[Int] = List(1, 2, 3, 4)
scala> List(1) match { case SortedSeq(a) => a }
res2: Int = 1
那么,您认为下面的示例展示了哪一个?
scala> List(1) match { case List(x) => x }
res3: Int = 1
一些例子:
scala> val fruit = List("apples", "oranges", "pears")
fruit: List[String] = List(apples, oranges, pears)
scala> val List(a, b, c) = fruit
a: String = apples
b: String = oranges
c: String = pears
scala> val List(a, b, _*) = fruit
a: String = apples
b: String = oranges
scala> val List(a, _*) = fruit
a: String = apples
scala> val List(a,rest @ _*) = fruit
a: String = apples
rest: Seq[String] = List(oranges, pears)
scala> val a::b::c::Nil = fruit
a: String = apples
b: String = oranges
c: String = pears
scala> val a::b::rest = fruit
a: String = apples
b: String = oranges
rest: List[String] = List(pears)
scala> val a::rest = fruit
a: String = apples
rest: List[String] = List(oranges, pears)
List(a,b,c)
可以被解释为匹配a :: b :: c :: Nil
,但是似乎 Scala 并不这样做;相反,它使用了List
的unapplySeq
方法。 - Dan Burton