如果這將被用於一些遊戲，我不會用數組，我會使用，因爲容易項目的列表可以添加/刪除。
Module Module1
Dim deck As New List(Of Integer)
Dim prng As New Random
Sub Main()
newDeck()
showDeck()
shuffle()
showDeck()
Dim s As String
Do
s = Console.ReadLine()
Select Case s
Case "s"
showDeck()
Case "c"
If deck.Count = 0 Then
Console.WriteLine("No cards")
Else
'take top card
Dim foo As Integer = deck(0)
deck.RemoveAt(0)
Console.WriteLine(foo.ToString)
End If
Case "n"
newDeck()
shuffle()
End Select
Loop Until s.ToLower = "x"
End Sub
Sub newDeck()
deck = Enumerable.Range(1, 52).ToList 'create deck
End Sub
Sub shuffle()
deck = deck.OrderBy(Function(r) prng.Next).ToList
End Sub
Sub showDeck()
Dim ctr As Integer = 0
Console.WriteLine()
For Each card As Integer In deck
Console.Write("{0}", card.ToString.PadLeft(4, " "c))
ctr += 1
If ctr Mod 10 = 0 Then Console.WriteLine()
Next
Console.WriteLine()
End Sub
End Module
一段時間以來，我一直認爲這是一個計算機不應該完全模仿現實世界的情況。首先，這是一個程序，用於瞭解隨機數生成器的種子的問題。
'create a form with three buttons and a richtextbox
Dim deckIdx As New List(Of Integer)
Dim cards As New List(Of card)
Dim prng As New Random
Private Sub Button1_Click(sender As System.Object, e As System.EventArgs) Handles Button1.Click
newDeck()
shuffle()
showDeck()
End Sub
Private Sub Button3_Click(sender As System.Object, e As System.EventArgs) Handles Button3.Click
'show next five cards
Dim sb As New System.Text.StringBuilder
sb.AppendLine()
For x As Integer = 1 To 5
Dim foo As card = getTopCard()
If Not foo Is Nothing Then sb.AppendFormat("{0}", foo.theCard.ToString.PadLeft(4, " "c))
Next
RichTextBox1.AppendText(sb.ToString)
RichTextBox1.ScrollToCaret()
End Sub
Private Sub Form1_Shown(sender As Object, e As System.EventArgs) Handles Me.Shown
Button1.PerformClick()
End Sub
Class card
Public theCard As Integer
Public count As Integer
End Class
Sub newDeck()
prng = New Random(42) '<<<<<<<<<<<<<<<<<<<<<<<< OOPS!!!
deckIdx = Enumerable.Range(0, 51).ToList 'create deck indicies
cards = New List(Of card)
For Each cIDX As Integer In deckIdx
Dim foo As New card
foo.theCard = cIDX
foo.count = 0
cards.Add(foo)
Next
End Sub
Sub shuffle()
deckIdx = deckIdx.OrderBy(Function(r) prng.Next).ToList
End Sub
Function getTopCard() As card
If deckIdx.Count > 0 Then
Dim foo As New card
foo.theCard = cards(deckIdx(0)).theCard
foo.count = cards(deckIdx(0)).count
deckIdx.RemoveAt(0)
Return foo
Else
Return Nothing
End If
End Function
Sub showDeck()
Dim ctr As Integer = 0
Dim sb As New System.Text.StringBuilder
For Each card As Integer In deckIdx
sb.AppendFormat("{0}", cards(card).theCard.ToString.PadLeft(4, " "c))
ctr += 1
If ctr Mod 10 = 0 Then sb.AppendLine()
Next
RichTextBox1.Text = sb.ToString
End Sub
運行該程序並重覆按下按鈕1/3將產生相同的結果，一遍又一遍。如果我們要堅持計算機必須完全模仿現實世界的想法，那麼顯而易見的解決方案是不重新實現PRNG。但是，如果我們採取了不同的方法。
像以前一樣添加該代碼
Dim bkgShuffle As Threading.Thread
Private Sub Button2_Click(sender As System.Object, e As System.EventArgs) Handles Button2.Click
Button2.Enabled = False
bkgShuffle = New Threading.Thread(AddressOf shuffleBkg)
bkgShuffle.IsBackground = True
bkgShuffle.Start()
End Sub
Sub shuffleBkg()
Do
Threading.Thread.Sleep(50)
shuffle()
Loop
End Sub
運行程序，按下按鈕1和3以確認一切都沒有改變。當您滿意時按下按鈕2，然後按下按鈕1和3.
當您親自玩牌時，顯然該牌不會不斷被洗牌，但如果它是？它會改變什麼嗎？如果我在線上玩牌，我是否知道或關心牌組是否在不斷洗牌？
這段代碼只是爲了證明一個觀點，在盒子外思考，而不應該被認爲是成品。
編輯： Bull Mountain可能會影響這一些。
可能重複/ 554587/is-there-an-easy-way-to-randomize-a-list-in-vb-net） –
@Bruno - 考慮使用列表。請參閱我提供的代碼。 – dbasnett