Question

我创建了一个以列表为参数的对象，但是当我将列表传递给执行对象时，出现错误。请指导，代码如下

object myobj {
    private var dateList  = List[String]()
    private val NPARAMS = 1

    private def parseArgs(args:Array[String]):Unit ={
        if(args.length < NPARAMS){
            println("Insufficient No of Parameters")
            System.exit(1)
        }
        var i = 0
        dateList = List(args(i))
    }
    // Some code...
}

正在执行对象

val dateList =  ("2019-01-01","2019-01-02")

myobj.main(Array(dateList))

错误

dateList: (String, String) = (2019-01-01,2019-01-02)

<console>:37: error: type mismatch;
 found   : (String, String)
 required: String
       myobj.main(Array(dateList))
                        ^

Answer 1

您的变量dateList是一个元组。在scala中，错误消息中的符号（TypeA，TypeB）是scala.Tuple2 [TypeA，TypeB]的简写。

您想要的是一个数组Array("arg1", "arg2")。

Answer 2

你是说

Outlook.Application app = new Outlook.Application();
RDOSession session = new RDOSession();
session.MAPIOBJECT = app.Session.MAPIOBJECT;
RDOFolder inbox = session.GetDefaultFolder(rdoDefaultFolders.olFolderInbox);
RDOItems items = inbox.Items;
RDOMail mail = items.GetFirst();
if (mail.MessageClass == "IPM.Note.SMIME") {
    RDOEncryptedMessage encryptedMessage = (RDOEncryptedMessage)session.GetMessageFromID(mail.EntryID)
    // from here I am stuck because encryptedMessage is null
}

？

如何在Scala中将List作为参数传递给对象

2 个答案: