Random.NextDouble() Method in C#

The Random.NextDouble() method in C# is used to return a random floating-point number that is greater than or equal to 0.0, and less than 1.0.

Syntax

The syntax is as follows −

public virtual double NextDouble ();

Example

Let us now see an example −

 Live Demo

using System;
public class Demo {
   public static void Main(){
      Random r1 = new Random();
      Random r2 = new Random();
      Byte[] arr = new Byte[2];
      r1.NextBytes(arr);
      Console.WriteLine("Random numbers in the byte array...");
      for (int i = 0; i < 2; i++)
         Console.WriteLine(arr[i]);
      Console.WriteLine("
Random floating point numbers...");
      for (int i = 0; i < 5; i++)
         Console.WriteLine(r2.NextDouble());
   }
}

Output

This will produce the following output −

Random numbers in the byte array...
124
141
Random floating point numbers...
0.93591266727816
0.36406785872023
0.122396959514542
0.795166163144245
0.954394097884369

Example

Let us now see another example −

 Live Demo

using System;
public class Demo {
   public static void Main(){
      int[] val = new int[7];
      Random r = new Random();
      double d;
      for (int i = 0; i 50; i++) {
         d = r.NextDouble();
         val[(int) Math.Ceiling(d*5)] ++;
      }
      Console.WriteLine("Random Numbers...");
      for (int i = 0; i < 7; i++)
         Console.WriteLine(val[i]);
   }
}

Output

This will produce the following output −

Random Numbers...
0
13
9
12
8
8
0

AmitDiwan

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Updated on: 03-Dec-2019

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