在测试夹具设置中运行时生成相同的随机数

发布于 2024-07-14 09:08:01 字数 242 浏览 7 评论 0原文

我试图在使用 NUnit 测试期间生成一个随机数，但它一直生成相同的数字。我为此目的使用以下函数。

dim dCount As Integer = Math.Floor((High - Low + 1) * Rnd() + Low)
dim divName As String = "abc" & dCount

知道为什么要这样做吗？

问候，

萨姆

原文

I am trying to generate a random number during testing using NUnit, but it keeps generating the same number. I am using the following function for this purpose.

dim dCount As Integer = Math.Floor((High - Low + 1) * Rnd() + Low)
dim divName As String = "abc" & dCount

Any idea why it is doing this?

Regards,

Sam

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苦笑流年记忆 2024-07-21 09:08:01

想必您正在快速连续执行许多测试。我不知道 Rnd() 在 VB 中具体做什么，但听起来它有典型的“每次调用新的 RNG”问题。

创建 Random 的单个实例并重复使用它。请注意，您的数学可以用一个简单的方法代替：

dim dCount as Integer = myRandom.Next(Low, High+1)

一个警告 - Random 不是线程安全的。如果您需要从不同线程生成随机数，要么使用锁定或线程静态。

另一点：使用随机数将使您的单元测试具有不确定性。你确定你必须这样做吗？有时它是合适的，但通常不是 IME。

Presumably you're executing many tests in quick succession. I don't know exactly what Rnd() does in VB, but it sounds like it's got the typical "new RNG per call" problem.

Create a single instance of Random and use it repeatedly. Note that your maths can be replaced by a simple:

dim dCount as Integer = myRandom.Next(Low, High+1)

One caveat - Random isn't thread-safe. If you need to generate random numbers from different threads, either use locking or thread statics.

On another point: using random numbers will make your unit tests non-deterministic. Are you sure you have to? Sometimes it's appropriate, but not often IME.

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林空鹿饮溪 2024-07-21 09:08:01

Dim dCount As Integer = between(low, high)
Dim divName As String = "abc" & dCount


Dim myRandom As New Random
Private Function between(ByVal low As Integer, ByVal high As Integer) As Integer
    between = myRandom.Next(low, high + 1)
End Function

Dim dCount As Integer = between(low, high)
Dim divName As String = "abc" & dCount


Dim myRandom As New Random
Private Function between(ByVal low As Integer, ByVal high As Integer) As Integer
    between = myRandom.Next(low, high + 1)
End Function

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