unapply 和 unapplySeq 有什么区别?

问题描述

为什么 Scala 有 unapply 和 unapplySeq?两者有什么区别?我什么时候应该更喜欢一个?

Why does Scala have both unapply and unapplySeq? What is the difference between the two? When should I prefer one over the other?

推荐答案

不深入细节和简化一点:

Without going into details and simplifying a bit:

对于常规参数 apply 构造和 unapply 解构:

For regular parameters apply constructs and unapply de-structures:

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 } 

对于重复参数，apply 构造和 unapplySeq 解构:

For repeated parameters, apply constructs and unapplySeq de-structures:

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，重复参数被视为 Seq 以及 A* 和 _* 之间的相似性.

Note that in that second case, repeated parameters are treated like a Seq and the similarity between A* and _*.

因此，如果您想解构自然包含各种单个值的内容，请使用 unapply.如果您想解构包含 Seq 的内容，请使用 unapplySeq.

So if you want to de-structure something that naturally contains various single values, use unapply. If you want to de-structure something that contains a Seq, use unapplySeq.

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