scala中是否存在用于从列表中删除项目的功能？

Posted 2021-04-27

我正在编写待办事项列表，我可以在其中添加元素并删除它们。我在尝试从可变列表中删除项目时遇到问题。

val list = MutableList[Any]()

def removeItem(){
     val input = scala.io.StdIn.readLine("
Enter item to remove:
").stripLineEnd

    for(i <- 0 to list.size){
        if (list[i] == input){
           list[i] = 0
           list -= input
        }
    }

}


def menu(option: Int): Boolean = {
  option match {
    case 1 =>
      addItem()
      true
    case 2 =>
      println("Enter item to remove")
      removeItem
      true
    case 3 =>
      println("
Here is your todo list:")
      display()
      true
    case 4 =>
      println("selected quit")
      false
    case _ => // the else case
      println("Unrecognized command")
      true
  }
}

[当我从菜单中调用该功能以删除项目时，我希望我的删除项目功能能够遍历可变列表并删除其中=用户输入的项目，并允许我每次执行此操作调用函数，但是我得到了这个错误：';'预期，但找到了'='。list [i] = 0

答案

您可以尝试使用scala.collection.mutable.ListBuffer。示例：

scala> import scala.collection.mutable.ListBuffer
import scala.collection.mutable.ListBuffer

scala> var mutableList = new ListBuffer[String]()
mutableList: scala.collection.mutable.ListBuffer[String] = ListBuffer()

scala> println(mutableList)// empty
ListBuffer()

scala> mutableList += "First element"
res1: scala.collection.mutable.ListBuffer[String] = ListBuffer(First element)

scala> println(mutableList)// 1 element
ListBuffer(First element)

scala> mutableList += "second element"
res3: scala.collection.mutable.ListBuffer[String] = ListBuffer(First element, second element)

scala> println(mutableList)// 2 elements
ListBuffer(First element, second element)

scala> // Deleting...

scala> mutableList -= "First element"
res5: scala.collection.mutable.ListBuffer[String] = ListBuffer(second element)