VB.NET Empty String Array

How can I create an empty one-dimensional string array?


Solution 1:

VB is 0-indexed in array declarations, so seomthing like Dim myArray(10) as String gives you 11 elements. It's a common mistake when translating from C languages.

So, for an empty array, either of the following would work:

Dim str(-1) as String ' -1 + 1 = 0, so this has 0 elements
Dim str() as String = New String() { } ' implicit size, initialized to empty

Solution 2:

Dim strEmpty(-1) As String

Solution 3:

The array you created by Dim s(0) As String IS NOT EMPTY

In VB.Net, the subscript you use in the array is index of the last element. VB.Net by default starts indexing at 0, so you have an array that already has one element.

You should instead try using System.Collections.Specialized.StringCollection or (even better) System.Collections.Generic.List(Of String). They amount to pretty much the same thing as an array of string, except they're loads better for adding and removing items. And let's be honest: you'll rarely create an empty string array without wanting to add at least one element to it.

If you really want an empty string array, declare it like this:

Dim s As String()

or

Dim t() As String

Solution 4:

Something like:

Dim myArray(9) as String

Would give you an array of 10 String references (each pointing to Nothing).

If you're not sure of the size at declaration time, you can declare a String array like this:

Dim myArray() as String

And then you can point it at a properly-sized array of Strings later:

ReDim myArray(9) as String

ZombieSheep is right about using a List if you don't know the total size and you need to dynamically populate it. In VB.NET that would be:

Dim myList as New List(Of String)
myList.Add("foo")
myList.Add("bar")

And then to get an array from that List:

myList.ToArray()

@Mark

Thanks for the correction.

Solution 5:

You don't have to include String twice, and you don't have to use New.
Either of the following will work...

Dim strings() as String = {}
Dim strings as String() = {}