How does C#'s random number generator work?

问题:

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 INTEGER numbers from 1-100.

回答1:

I was just wondering how the random number generator in C# works.

That's implementation-specific, but the wikipedia entry for pseudo-random number generators should give you some ideas.

I was also curious how I could make a program that generates random WHOLE INTEGER numbers from 1-100.

You can use 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.

You should also be aware of some of the "gotchas" associated with Random - in particular, you should not create 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 topic for more details.

回答2:

You can use 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.

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:

// 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3:

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.

{
  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,

1. create a 1 dimensional integer array.
2. full up the array with unordered numbers.
3. use the rnd.Next to get the position of the number that will be picked.

This works well.

To obtain a random number less than 100 use

{
  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.

Hope this assists someone positively.