CODEWITHSUNDEEP
c#
Jonas Bohez
Jonas Bohez

Reputation: 99

Random numbers every .Next() call [ Multiple .Next() in method]

I have a problem with random number generating in C#

I have tried all the solutions that I've found. Most we're only using the same random object every time which works only each object, not each property.

So I am using this as code for the random numbers:

private static readonly Random random = new Random();
private static readonly object syncLock = new object();

public static int RandomNumber(int min, int max)
{
            lock (syncLock)
            { // synchronize
                return random.Next(min, max);
            }
}

Now I want to call .Next() multiple times in the same method, to generate a random object:

public void StartDuiven()
{
   for (int i = 0; i <= 6; i++)
   {
      var d = new Duif();
      d.Naam = /*NameGenerator.GenRandomFirstName() +" "+ NameGenerator.GenRandomLastName()*/ "Jantje";
      d.GeboorteDatum = DateTime.Now;
      d.Snelheid = Genrator.RandomNumber(0, 4);
      d.Vliegtechniek = Genrator.RandomNumber(0, 4);
      d.Vormpeil = Genrator.RandomNumber(0, 4);
      d.Conditie = Genrator.RandomNumber(0, 4);
      d.Aerodynamica = Genrator.RandomNumber(0, 4);
      d.Intelligentie = Genrator.RandomNumber(0, 4);
      d.Libido = Genrator.RandomNumber(0, 4);
      d.Nachtvliegen = Genrator.RandomNumber(0, 4);
      d.Navigatie = Genrator.RandomNumber(0, 4);
      d.Ervaring = Genrator.RandomNumber(0, 4);
      d.Transfer = false;

      int g = Genrator.RandomNumber(0, 2); // Edited

      if (g == 0)
         d.Geslacht = false;
      else
         d.Geslacht = true;

      AddDuif(d);
   }
}

Every new object I get a different number, but not every time I call .Next() method.

So all the values of the properties become the same for 1 object.

How could I fix this? Why don't I get a new value on each .Next() call?

Thank you

Bye!

Upvotes: 4

Views: 196

Answers (3)

Alexei Levenkov
Alexei Levenkov

Reputation: 100547

Due to pigeonhole principle it is hard :) to get 5 distinct integer value in range of [0-4).

Since numbers are random you should expect that there will be some number of cases when all values will be the same.

(in addition to bug identified by @IngisKahn with usage of Random.Next(0,1))

Upvotes: 2

user1088520
user1088520

Reputation:

Try with 

return (random.Next() % (max - min)) + min

Upvotes: 0

IngisKahn
IngisKahn

Reputation: 869

Random.Next(0, 1) will always return 0. The max value is exclusive.

Upvotes: 4

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