# Random Int in C#

This article will introduce a method to generate a random integer number in C#.

## Use the `Next()` Method to Generate a Random Integer Number in `C#`

We will use the `Next()` method to generate a random integer number in C#. This method generates a random integer. It has three overloads. To use this method, we will have to create an object of the `Random` class. The correct syntax to use this method is as follows.

``````Random myObject = new Random();
myObject.Next();
``````

This method returns a random integer.

The program below shows how we can use the `Next()` method to generate a random integer.

``````using System;
public class Program {
public static void Main() {
Random myObject = new Random();
int ranNum= myObject.Next();
Console.WriteLine("The Random Number is: "+ranNum);
}
}
``````

Output:

``````The Random Number is: 880084995
``````

The random number generated by the function is too large. We can pass a number to this function so that it will return a random number less than that number.

The program below shows how we can use the `Next()` method to generate a random integer less than a specific number.

``````using System;
public class Program {
public static void Main() {
Random myObject = new Random();
int ranNum= myObject.Next(100);
Console.WriteLine("The Random Number is: "+ranNum);
}
}
``````

Output:

``````The Random Number is: 96
``````

We can also set a range. The function will return the random number between that range.

The program below shows how we can use the `Next()` method to generate a random integer within a range.

``````using System;
public class Program {
public static void Main() {
Random myObject = new Random();
int ranNum= myObject.Next(100, 150);
Console.WriteLine("The Random Number is: "+ranNum);
}
}
``````

Output:

``````The Random Number is: 145
``````

## Related Article - Csharp Integer

• C# Convert Int to String
• Convert Int to Enum in C#
• Random Number in a Range in C#
• Convert String to Int in C#
• Integer Division in C#