How do I generate a random int number?

Question

How do I generate a random integer in C#?

Answer 1

I wanted to add a cryptographically secure version:

RNGCryptoServiceProvider Class (MSDN or dotnetperls)

It implements IDisposable.

using (RNGCryptoServiceProvider rng = new RNGCryptoServiceProvider())
{
   byte[] randomNumber = new byte[4];//4 for int32
   rng.GetBytes(randomNumber);
   int value = BitConverter.ToInt32(randomNumber, 0);
}

Answer 2

I've tried all of these solutions excluding the COBOL answer... lol

None of these solutions were good enough. I needed randoms in a fast for int loop and I was getting tons of duplicate values even in very wide ranges. After settling for kind of random results far too long I decided to finally tackle this problem once and for all.

It's all about the seed.

I create a random integer by parsing out the non-digits from Guid, then I use that to instantiate my Random class.

public int GenerateRandom(int min, int max)
{
    var seed = Convert.ToInt32(Regex.Match(Guid.NewGuid().ToString(), @"\d+").Value);
    return new Random(seed).Next(min, max);
}

Update: Seeding isn't necessary if you instantiate the Random class once. So it'd be best to create a static class and call a method off that.

public static class IntUtil
{
   private static Random random;

   private static void Init()
   {
      if (random == null) random = new Random();
   }

   public static int Random(int min, int max)
   {
      Init();
      return random.Next(min, max);
   }
}

Then you can use the static class like so..

for(var i = 0; i < 1000; i++)
{
   int randomNumber = IntUtil.Random(1,100);
   Console.WriteLine(randomNumber); 
}

I admit I like this approach better.

Answer 3

Random r = new Random();
int n = r.Next();

Answer 4

create a Random object

Random rand = new Random();

and use it

int randomNumber = rand.Next(min, max);

you don't have to initialize new Random() every time you need a random number, initiate one Random then use it as many times as you need inside a loop or whatever

Answer 5

I used below code to have a random number (not recommended):

var random = new Random((int)DateTime.Now.Ticks);
var randomValue = random.Next(startValue, endValue + 1);

Answer 6

There are a number utility functions or services that are better cached in the same way that System.Random should be, so it lends itself to a generic implementation:

static public class CachedService<T> where T : new() {
    static public T Get { get; } = new T();
}

To use for random (or similar):

CachedService<System.Random>.Get.Next(999);