Why do I keep getting two of same random values in this code?

Question

Possible Duplicate:
Why does it appear that my random number generator isn't random in C#?

I have the following code:

        int a;
        int aa;

        Random aRand = new Random();
        Random aaRand = new Random();

        a = aRand.Next(20);
        aa = aaRand.Next(20);

        //if (a == aa)
        {
            Console.WriteLine(a + " " + aa);
            Console.ReadLine();
        }

I'm assuming that aRand and aaRand would be two different values, but that's not the case. What am I doing wrong? I'm assuming that aRand and aaRand will not always be the same, but they keep coming out the same all the time.

Thanks

c# correct is the language, sry. – Liger86 Oct 01 '11 at 22:20 — Liger86, Oct 01 '11 at 22:20

score 6 · Answer 1 · answered Oct 01 '11 at 22:22

This is explicitly covered in the docs for Random():

The default seed value is derived from the system clock and has finite resolution. As a result, different Random objects that are created in close succession by a call to the default constructor will have identical default seed values and, therefore, will produce identical sets of random numbers.

score 3 · Answer 2 · answered Oct 01 '11 at 22:21

Why are you creating two different Random variables? You could use just one:

int a;
int aa;

Random aRand = new Random();

a = aRand.Next(20);
aa = aRand.Next(20);

//if (a == aa)
{
    Console.WriteLine(a + " " + aa);
    Console.ReadLine();
}

Edit: "The random number generation starts from a seed value. If the same seed is used repeatedly, the same series of numbers is generated. One way to produce different sequences is to make the seed value time-dependent, thereby producing a different series with each new instance of Random. By default, the parameterless constructor of the Random class uses the system clock to generate its seed value, while its parameterized constructor can take an Int32 value based on the number of ticks in the current time. " from http://msdn.microsoft.com/en-us/library/system.random.aspx

score 1 · Answer 3 · answered Oct 01 '11 at 22:20

1

You only need one instance of Random() - just call .Next() twice.

int a;
int aa;

Random aRand = new Random();

a = aRand.Next(20);
aa = aRand.Next(20);

answered Oct 01 '11 at 22:20

Amber

507,862
82
626
550

score 1 · Answer 4 · answered Oct 01 '11 at 22:24

1

You should never have more than One Random variable in your entire application. Get rid of the second

Random aaRand = new Random();

answered Oct 01 '11 at 22:24

samack

815
11
28

score 1 · Answer 5 · answered Oct 01 '11 at 22:25

It looks like the two instances are using the same seed.

the seed determines all the values that will be generated and in which order. If you create 200 instances of Random with the same seed, they'll all give you the same output.

Create a single Instance when your app starts and reuse it.

Why do I keep getting two of same random values in this code?

5 Answers5

Linked

Related