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C# | Random.Next() Method

  • Last Updated : 30 Apr, 2019

The Next() Method of System.Random class in C# is used to get a random integer number. This method can be overloaded by passing different parameters to it as follows:

  • Next()
  • Next(Int32)
  • Next(Int32, Int32)

Next() Method

This method is used to returns a non-negative random integer.

Syntax:

public virtual int Next ();

Return Value: This method returns the 32-bit signed integer which is greater than or equal to 0 and less than MaxValue.



Example:




// C# program to illustrate the 
// Random.Next() Method
using System;
  
class GFG {
  
    // Driver code
    public static void Main()
    {
        // Instantiate random number generator
        Random rand = new Random();
  
        // Print 10 random numbers
        Console.WriteLine("Printing 10 random numbers");
        for (int i = 1; i <= 10; i++)
            Console.WriteLine("{0} -> {1}", i, rand.Next());
    }
}
Output:
Printing 10 random numbers
1 -> 1386420123
2 -> 2133003862
3 -> 981665925
4 -> 495382685
5 -> 1127976381
6 -> 824414652
7 -> 213006792
8 -> 1948108820
9 -> 214839986
10 -> 261560448

Next(Int32) Method

This method is used to get a non-negative random integer which is less than the specified maximum.

Syntax:

public virtual int Next (int maxValue);

Here, maxValue is the upper boundary of the random number to be generated. It must be greater than or equal to 0.

Return Value: The function returns a 32-bit signed integer which is greater than or equal to 0, and less than maxValue. However, if maxValue equals 0, maxValue is returned.

Exception: This method will give ArgumentOutOfRangeException if the maxValue is less than 0.

Example:






// C# program to illustrate the
// Random.Next(Int32) Method
using System;
  
class GFG {
  
    // Driver code
    public static void Main()
    {
        // Instantiate random number generator
        Random rand = new Random();
  
        // Print 10 random numbers less than 100
        Console.WriteLine("Printing 10 random numbers less than 100");
        for (int i = 1; i <= 10; i++)
            Console.WriteLine("{0} -> {1}", i, rand.Next(100));
    }
}
Output:
Printing 10 random numbers less than 100
1 -> 19
2 -> 94
3 -> 14
4 -> 54
5 -> 94
6 -> 73
7 -> 39
8 -> 42
9 -> 18
10 -> 77

Next(Int32, Int32) Method

This method is used to get the random integer that is within a specified range.

Syntax:

public virtual int Next (int minValue, int maxValue);

Parameters:

  • maxValue: It is the exclusive upper boundary of the random number generated. It must be greater than or equal to minValue.
  • minValue: It is the inclusive lower bound of the random number returned.

Return Value: The function returns a 32-bit signed integer greater than or equal to minValue and less than maxValue; that is, the range of return values includes minValue but not maxValue. If minValue equals maxValue, minValue is returned.

Exception: This method will give the ArgumentOutOfRangeException if the minValue is greater than maxValue.

Example:




// C# program to illustrate the
// Next(Int32, Int32) Method
using System;
  
class GFG {
  
    // Driver code
    public static void Main()
    {
        // Instantiate random number generator
        Random rand = new Random();
  
        // Print 10 random numbers between 50 and 100
        Console.WriteLine("Printing 10 random numbers"+
                                " between 50 and 100");
        for (int i = 1; i <= 10; i++)
            Console.WriteLine("{0} -> {1}", i, rand.Next(50, 100));
    }
}
Output:
Printing 10 random numbers between 50 and 100
1 -> 91
2 -> 85
3 -> 93
4 -> 74
5 -> 88
6 -> 77
7 -> 92
8 -> 76
9 -> 77
10 -> 52

Reference:




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