试着去了解我怎么能在Scala中使用类型:斯卡拉类型编译错误
object TypeSample extends App {
type MyParams = Map[Int, String]
def showParams(params: MyParams) = {
params.foreach(x => x match { case (a, b) => println(a + " " + b) })
}
//val params = MyParams(1 -> "one", 2 -> "two")
val params = Map(1 -> "one", 2 -> "two")
showParams(params)
}
此行抛出异常编译: “无法解析符号 'MyParams'”
//val params = MyParams(1 -> "one", 2 -> "two")
为什么?我不能像这样使用'type'?
还要注意的是,而不是'(X => X匹配{情况下(A,B)=> ...'你可以只写'{情况(a,b)=> ...' –