Scala案例类接受什么输入？

Question

The code below seems to work fine. But my question is that why the function is defined to take Strings as input but on the bottom, it has accepted some input integers?

case class

TempData(year:Int, month:String, Prec:Double, Maxtemp:Int, Meantemp:Int, Mintemp:Int)
    def parseLine(line:String):TempData = {
        val p = line.split(",")
        TempData(p(1).toInt, p(3).toString, p(5).toDouble, p(6).toInt, p(8).toInt, p(10).toInt)
    }

    //> parseLine: (line: String)Tests.TempData
    parseLine("Verizon, 2017, Alpha, October, gentlemen, 10.3, 5, Dallas, 67, schools, 42")
    //> res0: Tests.TempData = TempData(2017, October, 10.3, 5, 67, 42)

Answer 1

Your function takes one single String as an argument..

def parseLine(line: String): TempData

And one single String is exactly what is passed to it:

scala> "Verizon,2017,Alpha,October,gentlemen,10.3,5,Dallas,67,schools,42"
res0: String = Verizon,2017,Alpha,October,gentlemen,10.3,5,Dallas,67,schools,42

The numbers are just plain text and are a part of that String , there are no integers involved. It's the body of your function that splits the String and extracts Int s from it.

edit: regarding the title of your post: all of that has nothing to do with your TempData case class. It seems that you are confusing the call of the case class constructor with the call to parseLine .

Answer 2

The defined function parseLine(line: String) takes String type arguments, and you have passed it correctly. The confusion is that you are considering the numbers present in the string as integers(Int) which is incorrect. The numbers present in the string are of type String only because you have enclosed them in quotes. You can also reason out for using toInt for converting the the numbers from String to Int type.