Question

每次迭代后如何打印List的元素？

scala> def carry(c: Int, list: List[Int]):List[Int] = (c, list) match {
     |
     |   case (0, xs) => xs
     |
     |  case (1, Nil) => List(1)
     |
     |  case (1, x :: xs) => println(x,xs)((1 - x) :: carry(x, xs))
     |
     |   case (_, _) => throw new IllegalArgumentException("Invalid input!!!")
     |   }
<console>:18: error: Unit does not take parameters
        case (1, x :: xs) => println(x,xs)((1 - x) :: carry(x, xs))

Answer 1

您对println(x, xs)的调用返回了Unit类型，而您试图用参数(1 - x) :: carry(x, xs)调用返回值。也许您已经知道了。花括号在这里是您的朋友，因为它们创建了一个包含的表达式。解决方法如下：

case (1, x :: xs) => {println(x, xs); (1 - x) :: carry(x, xs)}

或

case (1, x :: xs) => {
  println(x, xs)
  (1 - x) :: carry(x, xs)
}

错误单位未接受参数scala

1 个答案: