How To Initialize A List Of String-Arrays

Question

Hello everyone,

I have a function f that takes a list of String-Arrays ( List(Of String()) ) as an argument.

Right now I'm passing it to the function like this:

Dim string1 As String() = {"foo", "bar"}
Dim string2 As String() = {"further", "words"}

Dim loS = New System.Collections.Generic.List(Of String()) From {string1, string2}

f(loS)

This seems unnecessarily complicated. But when I try to do it like this:

f( { {"foo", "bar"}, {"further", "words"} } )

I get an error like:

2-dimensional Array of String cannot be converted to System.Collections.Generic.List()

What am I doing wrong? How can I initialize an Object of type List(Of String())?

Thanks in davance

Answer 1

Hello,

Here are two methods to work with

Module Program  
    Sub Main(args As String())  
        Example1()  
        Console.WriteLine()  
        Example2()  
        Console.ReadLine()  
    End Sub  
    Private Sub Example1()  
        Dim myList As New List(Of String()) From {  
                New String() {"foo", "bar"},  
                New String() {"further", "words"}}  
  
        Console.WriteLine("Example1")  
  
        For Each strings As String() In myList  
            Console.WriteLine(String.Join(",", strings))  
        Next  
  
    End Sub  
  
    Private Sub Example2()  
  
        Dim string1 As String() = {"foo", "bar"}  
        Dim string2 As String() = {"further", "words"}  
  
        Dim myList As New List(Of String()) From {string1, string2}  
  
        Console.WriteLine("Example2")  
  
        For Each strings As String() In myList  
            Console.WriteLine(String.Join(",", strings))  
        Next  
    End Sub  
End Module  
  

Answer 2

Try a longer way:

f2( { ({"foo", "bar"}), ({"further", "words"}) } )

where f2 is a helper function:

Sub f2(p()() As String)
    f(p.ToList)
End Sub

Or, if possible, change the parameter type of f to call it directly.

Answer 3

Can confirm that both solutions work. Thank you! I picked the second one, because it's better suited for my particular problem. A helper function can however be a very elegant solution.