Scala - 如何从 java.io.Serializable 类型中提取列表

Question

I want to pass List along with other string parameters as argument. When I am trying to retrieve the list in called function it gives type Mismatch error.

   val list = ("abc","def","xyz")
    val ex_args = Array("--arg1","arg2",list)
    lst_args = ex_args.toList       

    // Get back the list in other function
    val mylist:List[String] = lst_args(2)   <------    This fails with type mismatch error 
ERROR:
        type mismatch;
        found   : java.io.Serializable
        required: List[String]

Answer 1

Okay, this will be difficult and painful for you, but I hope it will give you motivation to start learning Scala. First of all, you are not creating a list, but a tuple with 3 elements, you can read what is a tuple here . To create a list you must use the List keyword ( val list = List("abc", "def", "xyz") ), you can read how to do that here . Secondly, when you use List(index) you will get an element of your list, not a list, in your example you will get a String, but not a list String. I suggest you read this book and practice a little.

The code itself for your question:

    val list = List("abc", "def", "xyz")
    val ex_args = mutable.ArrayBuffer("--arg1", "arg2").addAll(list)
    val lst_args = ex_args
    
    val myList: List[String] = lst_args.toList

    val singleElement = myList(2)

    println(singleElement)