VB.Net Use string to construct name of array for use in function?

Question

I have two sets of jagged arrays, named P1 through P6 and Z1 through Z6, the contents of which are fruit in the P arrays and cars in the Z arrays. I also have a function that will obtain a value from one of these arrays:

Public Function retL(ByVal L As Array)
    Dim maxL As Integer = L.GetUpperBound(0)
    Dim numL As Integer = randomizer(maxL)
    Dim resL As String = L.GetValue(numL - 1)
    Return resL
End Function

L, in the above, is constructed by the following piece of code:

 Dim L As String = "P" & randomizer(6)
       or
 Dim L As String = "Z" & randomizer(6)

So, my issue is that none of this is working. Because L is constructed as a string, but then called by retL as an array, it fails. If I attempt to construct L as an array to begin with, it fails. If I attempt to call it as a string, it fails. I'm lost and confused as to where to proceed from here. Obviously what I'm trying to do is possible, but I've also obviously either missed an important step or I'm not grasping some basic concept.

Any suggestions?

Try not to think of an array as a data type in and of itself. Avoid using the keyword Array in your code. Instead, always look for an array of some specific type. — Joel Coehoorn
– Joel Coehoorn, Commented Oct 18, 2012 at 18:08

Joel Coehoorn · Accepted Answer · 2012-10-18 18:03:33Z

1

What you really want here is an array of arrays. Instead of:

Dim P1(n) As String
Dim P2(n) As String
'... 
Dim P6(n) As String
'...
Dim Z6(n) As String

... where 'n' is the size of your array, you should have something more like this:

Dim P(5, n) As String
Dim Z(5, n) As String

Note that I used a 5 instead of a 6, because in VB.Net arrays are zero indexed, rather than one indexed, and the subscript is the index of the last item, rather than the number of items. So Dim P(5, n) As String gives you a two-dimensional string array with six items on the first dimension.

Now your function should look like this:

Public Function retL(ByVal L() As String) As String
    Return L(randomizer(L.Length)-1)
End Function

And you could call the function from your current code like this:

Dim result As String = retL(P(randomizer(6) - 1)

edited Oct 18, 2012 at 18:03

answered Oct 18, 2012 at 17:58

Joel Coehoorn

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Comments

Olivier Jacot-Descombes · Accepted Answer · 2012-10-18 18:23:00Z

0

Here is a litle test class that does several things

Class TestRandomArrays
    Private L As String()
    Private randomizer As New Random()

    Public Sub InitializeArrayWithRandomLength()
        ' Creates an array of length between 0 and 5
        L = New String(randomizer.[Next](6) - 1) {}
    End Sub

    Public Sub FillArrayWithRandomText()
        For i As Integer = 0 To L.Length - 1
            L(i) = "P" & randomizer.[Next]()
        Next
    End Sub

    Public Function ReturnRandomArrayItem() As String
        Return L(randomizer.[Next](L.Length))
    End Function
End Class

Array items are accessed by their index using the array(index) syntax.

The brackets around the Next method are due to the fact that Next is a keyword in VB. They tell the VB compiler that Next is intended to be an identifier and not a keyword.

Random.Next(max) returns a integer between 0 and max-1.

New String(n) {} defines an array with an index range of 0..n and thus an array of length n+1.

This is NOT a jagged array. A jagged array is an array of arrays, i.e., the resulting variable would have (at least) two dimensions.

edited Oct 18, 2012 at 18:23

answered Oct 18, 2012 at 18:17

Olivier Jacot-Descombes

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3 Comments

user1757008 Over a year ago

Thanks to both of you for explaining. I did mis-speak in the opener of my question: The previous attempt at getting this item to work had been done with arrays of arrays, which is when I found out all about jagged arrays and how much VB.net apparently hates them and thinks no one should use them. I think between these two answers I have everything I need to get my project a little further down the road.

Olivier Jacot-Descombes Over a year ago

See my other post (especially the UPDATE).

Olivier Jacot-Descombes Over a year ago

It is hard to explain how to do it with jagged arrays, as I don't see anything in your code that could to be done with jagged arrays. Could you do what you have in mind with a two dimensional array instead? This would help us to understand your question better and to provide you a solution unsing jagged arrays. BTW: Your function should have a return type like Public Function retL(ByVal L As Array) As Integer. The purpose of the function would be clearer like this.

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