Question

为什么Set.empty与"".split(",").toSet中的PartialFunction行为不同？

val x = Set.empty
val y = Set("foo")
val z = "".split(",").toSet

val fooBar: PartialFunction[String, Int] = {
  case "foo" => 5
  case "bar" => 6
}

y.map(fooBar) // works
x.map(fooBar) // works
z.map(fooBar) // scala.MatchError:  (of class java.lang.String)

Answer 1

两者并不相同：

Set-Mailbox -identity s1112223 -Alias $SamAccountName ("{0}.{1}" -f $_.givenname, $_.sn)

scala> val x = Set.empty x: scala.collection.immutable.Set[Nothing] = Set() scala> x.size res0: Int = 0 scala> val z = "".split(",").toSet z: scala.collection.immutable.Set[String] = Set("") scala> z.size res1: Int = 1行生成一个由一个元素组成的集合，该元素是空字符串。您在输出中只看到split但实际上在Set()内是空字符串。：）

Set.empty上的匹配行为与＆＃34;＆＃34; .split（＆＃34;，＆＃34;）。toSet（PartialFunction）的行为不同

1 个答案: