unapply和unapplySeq有什么区别？

Question

为什么Scala同时有unapply和unapplySeq？两者之间的区别是什么？什么时候我应该更喜欢其中的一个呢？

Answer 1

不深入细节并稍微简化一下：

对于常规参数apply构造和unapply解构：

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构造和unapplySeq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，在第二种情况下，重复的参数被视为Seq以及A*和_*之间的相似性。

因此，如果您想要对自然包含各种单一值的内容进行解构，请使用unapply。如果想要解构包含Seq的内容，请使用unapplySeq。

Answer 2

固定性与可变性。Pattern Matching in Scala (pdf)通过镜像示例很好地解释了这一点。我在this answer中也有镜像的例子。

简要地说：

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，您认为下面的示例中展示了哪一个？

scala> List(1) match { case List(x) => x }
res3: Int = 1

Answer 3

下面是一些例子：

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)