Random.NextBytes() Method in C#


The Random.NextBytes() method in C# is used to fill the elements of a specified array of bytes with random numbers.

Syntax

The syntax is as follows −

public virtual void NextBytes (byte[] buffer);

Above the buffer is the array of bytes.

Example

Let us now see an example −

 Live Demo

using System;
public class Demo {
   public static void Main(){
      Random r = new Random();
      Random r2 = new Random();
      Random r3 = new Random();
      Byte[] arr = new Byte[5];
      r3.NextBytes(arr);
      Console.WriteLine("Random numbers.....");
      for (int i = 1; i <= 5; i++)
         Console.WriteLine(r.Next());
      Console.WriteLine("\nRandom numbers from 1 to 10.....");
      for (int i = 1; i <= 5; i++)
         Console.WriteLine(r2.Next(10));
      Console.WriteLine("\nRandom numbers in the byte array...");
      for (int i = 0; i < arr.GetUpperBound(0); i++)
         Console.WriteLine(arr[i]);
   }
}

Output

This will produce the following output −

Random numbers.....
2081486546
329484380
1639318640
1499756340
2122408387
Random numbers from 1 to 10.....
9
1
7
6
9
Random numbers in the byte array...
210
92
112
52

Example

Let us now see another example −

 Live Demo

using System;
public class Demo {
   public static void Main(){
      Random r = new Random();
      Byte[] arr = new Byte[2];
      r.NextBytes(arr);
      Console.WriteLine("Random numbers in the byte array...");
      for (int i = 0; i < 2; i++)
         Console.WriteLine(arr[i]);
   }
}

Output

This will produce the following output −

Random numbers in the byte array...
173
11
raja
Published on 03-Dec-2019 09:29:15
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