Question

如何将包含代码块的函数编码为包含case语句的参数？例如，在我的代码块中，我不想明确地进行匹配或默认情况。我看起来像这样

myApi {
    case Whatever() => // code for case 1
    case SomethingElse() => // code for case 2
}

在我的myApi（）中，它实际上会执行代码块并进行匹配。

Answer 1

您必须使用PartialFunction。

scala> def patternMatchWithPartialFunction(x: Any)(f: PartialFunction[Any, Unit]) = f(x)
patternMatchWithPartialFunction: (x: Any)(f: PartialFunction[Any,Unit])Unit

scala> patternMatchWithPartialFunction("hello") {
     |   case s: String => println("Found a string with value: " + s)
     |   case _ => println("Found something else")
     | }
Found a string with value: hello

scala> patternMatchWithPartialFunction(42) {
     |   case s: String => println("Found a string with value: " + s)
     |   case _ => println("Found something else")
     | }
Found something else

Answer 2

这应该足以解释它：A Tour of Scala: Pattern Matching

如何在scala中编写匹配代码块的模式？

2 个答案: