# 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
```