C# 的随机数生成器是如何工作的？
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How does C#'s random number generator work?
提问by Seth Taddiken
I was just wondering how the random number generator in C# works. I was also curious how I could make a program that generates random WHOLE INTEGERnumbers from 1-100.
我只是想知道 C# 中的随机数生成器是如何工作的。我也很好奇我如何制作一个程序来生成从 1 到 100 的随机整数。
采纳答案by Jon Skeet
I was just wondering how the random number generator in C# works.
我只是想知道 C# 中的随机数生成器是如何工作的。
That's implementation-specific, but the wikipedia entry for pseudo-random number generatorsshould give you some ideas.
这是特定于实现的，但伪随机数生成器的维基百科条目应该给你一些想法。
I was also curious how I could make a program that generates random WHOLE INTEGER numbers from 1-100.
我也很好奇我如何制作一个程序来生成从 1 到 100 的随机整数。
You can use Random.Next(int, int):
您可以使用Random.Next(int, int)：
Random rng = new Random();
for (int i = 0; i < 10; i++)
{
Console.WriteLine(rng.Next(1, 101));
}
Note that the upper bound is exclusive- which is why I've used 101 here.
请注意，上限是唯一的- 这就是我在这里使用 101 的原因。
You should also be aware of some of the "gotchas" associated with Random- in particular, you should notcreate a new instance every time you want to generate a random number, as otherwise if you generate lots of random numbers in a short space of time, you'll see a lot of repeats. See my article on this topicfor more details.
你也应该了解一些相关联的“陷阱”的Random-特别是，你应该不是创建一个新的实例要生成一个随机数每一次，否则，如果你在很短的时间空间产生大量的随机数，你会看到很多重复。有关更多详细信息，请参阅我关于此主题的文章。
回答by Zbigniew
You can use Random.Next(int maxValue):
您可以使用Random.Next(int maxValue)：
Return: A 32-bit signed integer greater than or equal to zero, and less than maxValue; that is, the range of return values ordinarily includes zero but not maxValue. However, if maxValue equals zero, maxValue is returned.
返回：一个大于或等于零且小于 maxValue 的 32 位有符号整数；也就是说，返回值的范围通常包括零但不包括maxValue。但是，如果 maxValue 等于零，则返回 maxValue。
var r = new Random();
// print random integer >= 0 and < 100
Console.WriteLine(r.Next(100));
For this case however you could use Random.Next(int minValue, int maxValue), like this:
但是，对于这种情况，您可以使用Random.Next(int minValue, int maxValue)，如下所示：
// print random integer >= 1 and < 101
Console.WriteLine(r.Next(1, 101);)
// or perhaps (if you have this specific case)
Console.WriteLine(r.Next(100) + 1);
回答by Arrai
I've been searching the internet for RNG for a while now. Everything I saw was either TOO complex or was just not what I was looking for. After reading a few articles I was able to come up with this simple code.
我已经在互联网上搜索 RNG 一段时间了。我看到的一切要么太复杂，要么不是我想要的。在阅读了几篇文章后，我能够想出这个简单的代码。
{
Random rnd = new Random(DateTime.Now.Millisecond);
int[] b = new int[10] { 5, 8, 1, 7, 3, 2, 9, 0, 4, 6 };
textBox1.Text = Convert.ToString(b[rnd.Next(10)])
}
Simple explanation,
简单的解释，
- create a 1 dimensional integer array.
- full up the array with unordered numbers.
- use the rnd.Next to get the position of the number that will be picked.
- 创建一个一维整数数组。
- 用无序数字填满数组。
- 使用 rnd.Next 获取将被选取的数字的位置。
This works well.
这很好用。
To obtain a random number less than 100 use
要获得小于 100 的随机数，请使用
{
Random rnd = new Random(DateTime.Now.Millisecond);
int[] b = new int[10] { 5, 8, 1, 7, 3, 2, 9, 0, 4, 6 };
int[] d = new int[10] { 9, 4, 7, 2, 8, 0, 5, 1, 3, 4 };
textBox1.Text = Convert.ToString(b[rnd.Next(10)]) + Convert.ToString(d[rnd.Next(10)]);
}
and so on for 3, 4, 5, and 6 ... digit random numbers.
依此类推 3、4、5 和 6 ... 位随机数。
Hope this assists someone positively.
希望这对某人有积极的帮助。
