Weird Random behavior

Question

Consider the following:

SomeObject o1 = new SomeObject("obj 1");
SomeObject o2 = new SomeObject("obj 2");

for (int i = 0; i < 10; i++)
{
     o1.Method();
     o2.Method();
}

Where:

public class SomeObject
{
    string name;
    public SomeObject(string name)
    {
        this.name = name;
    }
    public void Method()
    {
        Console.WriteLine($"{name}:  {new Random().Next(1, 101)}");
    }
}

Output:

obj 1: 99

obj 2: 99

obj 1: 99

obj 2: 99

obj 1: 99

obj 2: 99

obj 1: 99

obj 2: 99

obj 1: 99

obj 2: 99

That's doesn't looks like random numbers

They are random, just not the way you’d like them to be. If you create new instances of Random they’ll get seeded similarly and will give same numbers out of them. — Sami Kuhmonen, May 02 '19 at 11:39

score 1 · Accepted Answer · answered May 02 '19 at 11:39

You need to create the Random once, and then use it many times.

Like this:

SomeObject o1 = new SomeObject("obj 1");
SomeObject o2 = new SomeObject("obj 2");
var random = new Random();

for (int i = 0; i < 10; i++)
{
    o1.Method(random);
    o2.Method(random);
}

With this class:

public class SomeObject
{
    string name;
    public SomeObject(string name)
    {
        this.name = name;
    }

    public void Method(Random random)
    {
        Console.WriteLine($"{name}:  {random.Next(1, 101)}");
    }
}

Or you could pass the Random instance in the constructor or something. The important thing is: the numbers are only "random" within the same instance.

score 0 · Answer 2 · answered May 02 '19 at 11:43

0

If the same seed is used for separate Random objects, they will generate the same series of random numbers.

Seed is the starting value.

Docs

answered May 02 '19 at 11:43

Shad

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I didn't do it. – Shad May 02 '19 at 12:47

Weird Random behavior

2 Answers2