How to Generate a Random Integer in C#

How do I generate a random integer in C#?

The Random class is used to create random numbers. (Pseudo-random that is of course.).

Example:

Random rnd = new Random();
int month  = rnd.Next(1, 13);  // creates a number between 1 and 12
int dice   = rnd.Next(1, 7);   // creates a number between 1 and 6
int card   = rnd.Next(52);     // creates a number between 0 and 51

If you are going to create more than one random number, you should keep the Random instance and reuse it. If you create new instances too close in time, they will produce the same series of random numbers as the random generator is seeded from the system clock.

Produce a random number in a range using C#

You can try

//for integers
Random r = new Random();
int rInt = r.Next(0, 100);

//for doubles
int range = 100;
double rDouble = r.NextDouble()* range;

Have a look at

Random Class, Random.Next Method (Int32, Int32) and Random.NextDouble Method

How to generate random numbers from a range in c#

You need this overload of Random.Next():

public virtual int Next(
    int minValue,
    int maxValue
)

Where:

minValue = The inclusive lower bound of the random number returned.

maxValue = The exclusive upper bound of the random number returned.
maxValue must be greater than or equal to minValue.

Note the words inclusive and exclusive in the parameter descriptions. This means the minimum value can be returned in the possible values, while the maximum value will not be possible.

This is further clarified in the description of the return value:

Return Value - 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.

For example, to get single digit values between 0 and 9 (inclusive), you'd use:

int value = rnd.Next(0, 10); // return a value between 0 and 9 inclusive

To get double digits, you'd use:

int value = rnd.Next(10, 100); // return a value between 10 and 99 inclusive

Finally, to get triple digit numbers, you'd use:

int value = rnd.Next(100, 1000); // return a value between 100 and 999 inclusive

Quickly generating random numbers in C#

Do not use multiple instances of the Random Class using the default constructor. If they all are initialized within the same time slice they will all have the same seed and will all generate the same sequence of random numbers, use the constructor that you can pass in a seed and pass a different seed to each instance.

Random rand0, rand1, rand2;

void init()
{
      int baseSeed = (int) DateTime.Now.Ticks;
      rand0 = new Random(baseSeed);
      rand1 = new Random(baseSeed + 1);
      rand2 = new Random(baseSeed + 2);
}

Also you only need one object per thread, use the same Random object for all of the panels if they are all on the same thread.

How do I generate random numbers x times?

var x = 10;
List<int> numbers = new List<int>();
Random random = new Random();
for (int i = 0; i < x; i++)
{
    numbers.Add(random.Next(1, 100));
}

Console.WriteLine(string.Join(", ", numbers));

Sample Image

Generating random numbers without repeating.C#

Check each number that you generate against the previous numbers:

List<int> listNumbers = new List<int>();
int number;
for (int i = 0; i < 6; i++)
{
  do {
     number = rand.Next(1, 49);
  } while (listNumbers.Contains(number));
  listNumbers.Add(number);
}

Another approach is to create a list of possible numbers, and remove numbers that you pick from the list:

List<int> possible = Enumerable.Range(1, 48).ToList();
List<int> listNumbers = new List<int>();
for (int i = 0; i < 6; i++)
{
  int index = rand.Next(0, possible.Count);
  listNumbers.Add(possible[index]);
  possible.RemoveAt(index);
}