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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 −
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 −
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("
Random 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
Updated on: 03-Dec-2019
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