Random.Next() Method in C#


The Random.Next() method in C# is used to return a non-negative random integer.

Syntax

The syntax is as follows −

public virtual int Next ();
public virtual int Next (int maxVal);

Above, the maxVal parameter is the exclusive upper bound of the random number to be generated.

Example

Let us now see an example −

 Live Demo

using System;
public class Demo {
   public static void Main(){
      Random r = new Random();
      Console.WriteLine("Random numbers.....");
      for (int i = 1; i <= 5; i++)
      Console.WriteLine(r.Next());
   }
}

Output

This will produce the following output −

Random numbers.....
1014639030
1510161246
1783253715
487417801
249480649

Example

Let us now see another example −

 Live Demo

using System;
public class Demo {
   public static void Main(){
      Random r = new Random();
      Random r2 = new Random();
      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));
   }
}

Output

This will produce the following output −

Random numbers.....
613432308
1705125884
1787561614
1243383842
2016323534
Random numbers from 1 to 10.....
2
7
8
5
9
raja
Published on 03-Dec-2019 09:53:14
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