Scala - 案例序列

Question

我只是将案例序列作为部分函数阅读,语法有点奇怪.

例如

def test: Int => Int = {
  case 1 => 2
  case 2 => 3
  case _ => 0
}    

我希望它test没有参数,并返回一个类型的函数Int => Int

但经过一些测试后,它似乎需要一个int作为参数并返回一个int,所以我把它重写为......

def test1(i: Int): Int =
  i match {
    case 1 => 2
    case 2 => 3
    case _ => 0
  }

是test和test1平等？

Answer 1

他们不平等.test是返回的方法Function1[Int,Int]和test1是需要一个的方法Int,并返回Int.这也与模式匹配表达完全无关.

Answer 2

前面的代码确实返回Int => Int类型的函数.

Welcome to Scala version 2.9.1.final (Java HotSpot(TM) Client VM, Java 1.6.0_25).
Type in expressions to have them evaluated.
Type :help for more information.

scala> :paste
// Entering paste mode (ctrl-D to finish)

def test: Int => Int = {
case 1 => 2
case 2 => 3
case _ => 0
}

// Exiting paste mode, now interpreting.

test: Int => Int

scala> test
res0: Int => Int = <function1>

scala> test.apply(1)
res1: Int = 2

也许令人困惑的是,可以直接调用apply:

scala> test(1)
res2: Int = 2