VB.NET 中的随机整数

StackOverflow https://stackoverflow.com/questions/18676

  •  09-06-2019
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题

我需要生成一个 1 到 n 之间的随机整数（其中 n 是正整数）用于单元测试。我不需要过于复杂的东西来确保真正的随机性 - 只需一个老式的随机数。

我该怎么做呢？

有帮助吗？

解决方案

要获得 1 到 N（含）之间的随机整数值，可以使用以下命令。

CInt(Math.Ceiling(Rnd() * n)) + 1

其他提示

正如多次指出的，这样编写代码的建议是有问题的：

Public Function GetRandom(ByVal Min As Integer, ByVal Max As Integer) As Integer
    Dim Generator As System.Random = New System.Random()
    Return Generator.Next(Min, Max)
End Function

原因是构造函数 Random 类提供基于系统时钟的默认种子。在大多数系统上，这具有有限的粒度——大约 20 毫秒。因此，如果您编写以下代码，您将连续多次获得相同的数字：

Dim randoms(1000) As Integer
For i As Integer = 0 to randoms.Length - 1
    randoms(i) = GetRandom(1, 100)
Next

下面的代码解决了这个问题：

Public Function GetRandom(ByVal Min As Integer, ByVal Max As Integer) As Integer
    ' by making Generator static, we preserve the same instance '
    ' (i.e., do not create new instances with the same seed over and over) '
    ' between calls '
    Static Generator As System.Random = New System.Random()
    Return Generator.Next(Min, Max)
End Function

我使用这两种方法编写了一个简单的程序来生成 1 到 100 之间的 25 个随机整数。这是输出：

Non-static: 70 Static: 70
Non-static: 70 Static: 46
Non-static: 70 Static: 58
Non-static: 70 Static: 19
Non-static: 70 Static: 79
Non-static: 70 Static: 24
Non-static: 70 Static: 14
Non-static: 70 Static: 46
Non-static: 70 Static: 82
Non-static: 70 Static: 31
Non-static: 70 Static: 25
Non-static: 70 Static: 8
Non-static: 70 Static: 76
Non-static: 70 Static: 74
Non-static: 70 Static: 84
Non-static: 70 Static: 39
Non-static: 70 Static: 30
Non-static: 70 Static: 55
Non-static: 70 Static: 49
Non-static: 70 Static: 21
Non-static: 70 Static: 99
Non-static: 70 Static: 15
Non-static: 70 Static: 83
Non-static: 70 Static: 26
Non-static: 70 Static: 16
Non-static: 70 Static: 75

使用 系统随机:

Dim MyMin As Integer = 1, MyMax As Integer = 5, My1stRandomNumber As Integer, My2ndRandomNumber As Integer

' Create a random number generator
Dim Generator As System.Random = New System.Random()

' Get a random number >= MyMin and <= MyMax
My1stRandomNumber = Generator.Next(MyMin, MyMax + 1) ' Note: Next function returns numbers _less than_ max, so pass in max + 1 to include max as a possible value

' Get another random number (don't create a new generator, use the same one)
My2ndRandomNumber = Generator.Next(MyMin, MyMax + 1)

到目前为止，所有答案都有问题或错误（复数，而不是一个）。我会解释。但首先我想赞扬丹涛的洞察力，使用静态变量来记住 Generator 变量，因此多次调用它不会一遍又一遍地重复相同的#，而且他给出了非常好的解释。但正如我现在所解释的，他的代码与大多数其他代码都存在同样的缺陷。

MS 的 Next() 方法相当奇怪。正如人们所期望的那样，Min 参数是包含的最小值，但 Max 参数是 独家的 最大程度出人意料。换句话说，如果您传递 min=1 和 max=5，那么您的随机数将是 1、2、3 或 4 中的任何一个，但绝不会包括 5。这是使用 Microsoft Random.Next() 方法的所有代码中两个潜在错误中的第一个。

为一个 简单的 答案（但仍然存在其他可能但罕见的问题）那么您需要使用：

Private Function GenRandomInt(min As Int32, max As Int32) As Int32
    Static staticRandomGenerator As New System.Random
    Return staticRandomGenerator.Next(min, max + 1)
End Function

（我喜欢用 Int32 而不是 Integer 因为它可以更清楚地显示 int 有多大，而且输入起来更短，但适合你自己。）

我发现此方法有两个潜在问题，但它适合（并且正确）大多数用途。所以如果你想要一个 简单的 解决方案，我相信这是正确的。

我发现这个函数唯一的两个问题是：1：当 Max = Int32.MaxValue 时，加 1 会导致数字溢出。尽管这种情况很少见，但仍然有可能。2：当最小值 > 最大值 + 1 时。当 min = 10 且 max = 5 时，Next 函数会抛出错误。这可能就是你想要的。但也可能不是。或者考虑当 min = 5 且 max = 4 时。通过添加 1，5 被传递给 Next 方法，但它不会抛出错误，当它确实是一个错误时，但我测试的 Microsoft .NET 代码返回 5。因此，当最大值 = 最小值时，它确实不是“排他”最大值。但是当 Random.Next() 函数的 max < min 时，它会抛出 ArgumentOutOfRangeException。所以微软的实现在这方面确实不一致并且有错误。

您可能只想在 min > max 时交换数字，这样就不会引发错误，但这完全取决于所需的内容。如果您希望在无效值上出现错误，那么最好在代码中 Microsoft 的独占最大值 (max + 1) 等于最小值时也抛出错误，在这种情况下，MS 不会出错。

处理 max = Int32.MaxValue 时的解决方法有点不方便，但我希望发布一个完整的函数来处理这两种情况。如果您想要与我编码方式不同的行为，请适合您自己。但要注意这两个问题。

快乐编码！

编辑：所以我需要一个随机整数生成器，我决定对其进行“正确”编码。因此，如果有人想要完整的功能，这里有一个真正有效的功能。（但它并没有赢得仅用 2 行代码的最简单的奖品。但它也并不复杂。）

''' <summary>
''' Generates a random Integer with any (inclusive) minimum or (inclusive) maximum values, with full range of Int32 values.
''' </summary>
''' <param name="inMin">Inclusive Minimum value. Lowest possible return value.</param>
''' <param name="inMax">Inclusive Maximum value. Highest possible return value.</param>
''' <returns></returns>
''' <remarks></remarks>
Private Function GenRandomInt(inMin As Int32, inMax As Int32) As Int32
    Static staticRandomGenerator As New System.Random
    If inMin > inMax Then Dim t = inMin : inMin = inMax : inMax = t
    If inMax < Int32.MaxValue Then Return staticRandomGenerator.Next(inMin, inMax + 1)
    ' now max = Int32.MaxValue, so we need to work around Microsoft's quirk of an exclusive max parameter.
    If inMin > Int32.MinValue Then Return staticRandomGenerator.Next(inMin - 1, inMax) + 1 ' okay, this was the easy one.
    ' now min and max give full range of integer, but Random.Next() does not give us an option for the full range of integer.
    ' so we need to use Random.NextBytes() to give us 4 random bytes, then convert that to our random int.
    Dim bytes(3) As Byte ' 4 bytes, 0 to 3
    staticRandomGenerator.NextBytes(bytes) ' 4 random bytes
    Return BitConverter.ToInt32(bytes, 0) ' return bytes converted to a random Int32
End Function

Public Function RandomNumber(ByVal n As Integer) As Integer
    'initialize random number generator
    Dim r As New Random(System.DateTime.Now.Millisecond)
    Return r.Next(1, n)
End Function

微软示例 Rnd 函数

https://msdn.microsoft.com/en-us/library/f7s023d2%28v=vs.90%29.aspx

1- 初始化随机数生成器。

Randomize()

2 - 生成 1 到 6 之间的随机值。 

Dim value As Integer = CInt(Int((6 * Rnd()) + 1))

如果您使用约瑟夫的答案，这是一个很好的答案，并且您像这样背靠背运行这些答案：

dim i = GetRandom(1, 1715)
dim o = GetRandom(1, 1715)

然后结果可能会一遍又一遍地返回相同的结果，因为它处理调用的速度非常快。这在 08 年可能不是问题，但由于现在的处理器速度更快，该函数不允许系统时钟在进行第二次调用之前有足够的时间进行更改。

由于 System.Random() 函数基于系统时钟，因此我们需要留出足够的时间让它在下一次调用之前发生变化。实现此目的的一种方法是将当前线程暂停 1 毫秒。请参阅下面的示例：

Public Function GetRandom(ByVal min as Integer, ByVal max as Integer) as Integer
    Static staticRandomGenerator As New System.Random
    max += 1
    Return staticRandomGenerator.Next(If(min > max, max, min), If(min > max, min, max))
End Function

您应该只创建一次伪随机数生成器：

Dim Generator As System.Random = New System.Random()

然后，如果整数足以满足您的需求，您可以使用：

Public Function GetRandom(myGenerator As System.Random, ByVal Min As Integer, ByVal Max As Integer) As Integer
'min is inclusive, max is exclusive (dah!)
Return myGenerator.Next(Min, Max + 1)
End Function

想要多少次都可以。使用包装函数是合理的，只是因为最大值是排他的 - 我知道随机数以这种方式工作，但 .Next 的定义令人困惑。

在我看来，每次需要一个数字时就创建一个生成器是错误的；伪随机数不是这样工作的。

首先，您遇到了初始化问题，这已在其他回复中讨论过。如果初始化一次，就不会出现这个问题。

其次，我完全不确定您是否获得了有效的随机数序列；相反，您会获得多个不同序列的第一个集合，这些序列是根据计算机时间自动播种的。我不确定这些数字是否会通过确认序列随机性的测试。

Dim rnd As Random = New Random
rnd.Next(n)

仅供参考，VB NET Fuction 对 RND 和 RANDOMIZE 的定义（应该给出与 BASIC（1980 年）和之后所有版本相同的结果）：

Public NotInheritable Class VBMath
    ' Methods
    Private Shared Function GetTimer() As Single
        Dim now As DateTime = DateTime.Now
        Return CSng((((((60 * now.Hour) + now.Minute) * 60) + now.Second) + (CDbl(now.Millisecond) / 1000)))
    End Function

    Public Shared Sub Randomize()
        Dim timer As Single = VBMath.GetTimer
        Dim projectData As ProjectData = ProjectData.GetProjectData
        Dim rndSeed As Integer = projectData.m_rndSeed
        Dim num3 As Integer = BitConverter.ToInt32(BitConverter.GetBytes(timer), 0)
        num3 = (((num3 And &HFFFF) Xor (num3 >> &H10)) << 8)
        rndSeed = ((rndSeed And -16776961) Or num3)
        projectData.m_rndSeed = rndSeed
    End Sub

    Public Shared Sub Randomize(ByVal Number As Double)
        Dim num2 As Integer
        Dim projectData As ProjectData = ProjectData.GetProjectData
        Dim rndSeed As Integer = projectData.m_rndSeed
        If BitConverter.IsLittleEndian Then
            num2 = BitConverter.ToInt32(BitConverter.GetBytes(Number), 4)
        Else
            num2 = BitConverter.ToInt32(BitConverter.GetBytes(Number), 0)
        End If
        num2 = (((num2 And &HFFFF) Xor (num2 >> &H10)) << 8)
        rndSeed = ((rndSeed And -16776961) Or num2)
        projectData.m_rndSeed = rndSeed
    End Sub

    Public Shared Function Rnd() As Single
        Return VBMath.Rnd(1!)
    End Function

    Public Shared Function Rnd(ByVal Number As Single) As Single
        Dim projectData As ProjectData = ProjectData.GetProjectData
        Dim rndSeed As Integer = projectData.m_rndSeed
        If (Number <> 0) Then
            If (Number < 0) Then
                Dim num1 As UInt64 = (BitConverter.ToInt32(BitConverter.GetBytes(Number), 0) And &HFFFFFFFF)
                rndSeed = CInt(((num1 + (num1 >> &H18)) And CULng(&HFFFFFF)))
            End If
            rndSeed = CInt((((rndSeed * &H43FD43FD) + &HC39EC3) And &HFFFFFF))
        End If
        projectData.m_rndSeed = rndSeed
        Return (CSng(rndSeed) / 1.677722E+07!)
    End Function

End Class

而随机类别是：

Public Class Random
    ' Methods
    <__DynamicallyInvokable> _
    Public Sub New()
        Me.New(Environment.TickCount)
    End Sub

    <__DynamicallyInvokable> _
    Public Sub New(ByVal Seed As Integer)
        Me.SeedArray = New Integer(&H38  - 1) {}
        Dim num4 As Integer = If((Seed = -2147483648), &H7FFFFFFF, Math.Abs(Seed))
        Dim num2 As Integer = (&H9A4EC86 - num4)
        Me.SeedArray(&H37) = num2
        Dim num3 As Integer = 1
        Dim i As Integer
        For i = 1 To &H37 - 1
            Dim index As Integer = ((&H15 * i) Mod &H37)
            Me.SeedArray(index) = num3
            num3 = (num2 - num3)
            If (num3 < 0) Then
                num3 = (num3 + &H7FFFFFFF)
            End If
            num2 = Me.SeedArray(index)
        Next i
        Dim j As Integer
        For j = 1 To 5 - 1
            Dim k As Integer
            For k = 1 To &H38 - 1
                Me.SeedArray(k) = (Me.SeedArray(k) - Me.SeedArray((1 + ((k + 30) Mod &H37))))
                If (Me.SeedArray(k) < 0) Then
                    Me.SeedArray(k) = (Me.SeedArray(k) + &H7FFFFFFF)
                End If
            Next k
        Next j
        Me.inext = 0
        Me.inextp = &H15
        Seed = 1
    End Sub

    Private Function GetSampleForLargeRange() As Double
        Dim num As Integer = Me.InternalSample
        If ((Me.InternalSample Mod 2) = 0) Then
            num = -num
        End If
        Dim num2 As Double = num
        num2 = (num2 + 2147483646)
        Return (num2 / 4294967293)
    End Function

    Private Function InternalSample() As Integer
        Dim inext As Integer = Me.inext
        Dim inextp As Integer = Me.inextp
        If (++inext >= &H38) Then
            inext = 1
        End If
        If (++inextp >= &H38) Then
            inextp = 1
        End If
        Dim num As Integer = (Me.SeedArray(inext) - Me.SeedArray(inextp))
        If (num = &H7FFFFFFF) Then
            num -= 1
        End If
        If (num < 0) Then
            num = (num + &H7FFFFFFF)
        End If
        Me.SeedArray(inext) = num
        Me.inext = inext
        Me.inextp = inextp
        Return num
    End Function

    <__DynamicallyInvokable> _
    Public Overridable Function [Next]() As Integer
        Return Me.InternalSample
    End Function

    <__DynamicallyInvokable> _
    Public Overridable Function [Next](ByVal maxValue As Integer) As Integer
        If (maxValue < 0) Then
            Dim values As Object() = New Object() { "maxValue" }
            Throw New ArgumentOutOfRangeException("maxValue", Environment.GetResourceString("ArgumentOutOfRange_MustBePositive", values))
        End If
        Return CInt((Me.Sample * maxValue))
    End Function

    <__DynamicallyInvokable> _
    Public Overridable Function [Next](ByVal minValue As Integer, ByVal maxValue As Integer) As Integer
        If (minValue > maxValue) Then
            Dim values As Object() = New Object() { "minValue", "maxValue" }
            Throw New ArgumentOutOfRangeException("minValue", Environment.GetResourceString("Argument_MinMaxValue", values))
        End If
        Dim num As Long = (maxValue - minValue)
        If (num <= &H7FFFFFFF) Then
            Return (CInt((Me.Sample * num)) + minValue)
        End If
        Return (CInt(CLng((Me.GetSampleForLargeRange * num))) + minValue)
    End Function

    <__DynamicallyInvokable> _
    Public Overridable Sub NextBytes(ByVal buffer As Byte())
        If (buffer Is Nothing) Then
            Throw New ArgumentNullException("buffer")
        End If
        Dim i As Integer
        For i = 0 To buffer.Length - 1
            buffer(i) = CByte((Me.InternalSample Mod &H100))
        Next i
    End Sub

    <__DynamicallyInvokable> _
    Public Overridable Function NextDouble() As Double
        Return Me.Sample
    End Function

    <__DynamicallyInvokable> _
    Protected Overridable Function Sample() As Double
        Return (Me.InternalSample * 4.6566128752457969E-10)
    End Function


    ' Fields
    Private inext As Integer
    Private inextp As Integer
    Private Const MBIG As Integer = &H7FFFFFFF
    Private Const MSEED As Integer = &H9A4EC86
    Private Const MZ As Integer = 0
    Private SeedArray As Integer()
End Class

Function xrand() As Long
        Dim r1 As Long = Now.Day & Now.Month & Now.Year & Now.Hour & Now.Minute & Now.Second & Now.Millisecond
        Dim RAND As Long = Math.Max(r1, r1 * 2)
        Return RAND
End Function

bboyse]这是最好的方法，从头开始：P

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