How random function inside random class generates an integer?

Question

I have an array of strings that I want to pick one variable randomly from the list. Although I am using random function but I would like to know whether it will pick value zero ever or not? (Element zero from my array)

string randomString = a[new Random().Next(a.Length)];

Did you read the [documentation](http://msdn.microsoft.com/en-us/library/system.random.aspx) first? — Soner Gönül, Feb 03 '14 at 13:30
The MSDN documentation of the `Random` class is very helpful. — Dirk, Feb 03 '14 at 13:31
"string randomString = a[new Random().Next(a.Length - 1)];" uhhh, what? — Weyland Yutani, Feb 03 '14 at 13:39
[`var index = new Random(0).Next() - 1559595546;`](http://dotnetfiddle.net/sZfrzS) will produce zero index. Whoever want it. For whatever reason. — Sinatr, Feb 03 '14 at 14:13

score 3 · Accepted Answer · answered Feb 03 '14 at 13:36

3

Random should be random, so getting an A should be Random

One thing to note is if you're calling

string randomString = a[new Random().Next(a.Length)];

in quick succession e.g. in a loop, then create a class field, else Next() will be the same.

private Random _random = new Random();

then in your function call

string randomString = a[_random.Next(a.Length)];

answered Feb 03 '14 at 13:36

ASG

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9
20

does random function generate value between 0 and 1? then it multiply it by the number of the elements of my array? – CowBoy Feb 03 '14 at 16:53
Next() only produces integers. – ASG Feb 03 '14 at 18:02

Winderps · Answer 2 · 2014-02-03T14:21:24.490

2

The Random class simply uses pseudorandom number generation to give you a number that fits your specifications and seems random. Your code is fine, but with (in theory) 26 possible results, getting one of them shouldn't occur too often.

I believe the issue you're experiencing is that you're creating new Random objects in a loop, and they're ending up with the same seed, try declaring it in your class:

Random rnd = new Random();

And using it like so:

string randomString = a[rnd.Next(a.Length)];

The benefit of using this (as opposed to your current method) is that if you run it inside a loop you don't risk getting the same answer multiple times in a row.

Hope this helped, cheers!

~Winderps

edited Feb 03 '14 at 14:21

answered Feb 03 '14 at 13:43

Winderps

100
9

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"Your code is fine, but with (in theory) 26 possible results, getting one of them shouldn't occur too often.". If only we knew how often.. – Weyland Yutani Feb 03 '14 at 13:45
"new Random().Next(a.Length-1)"? If you want to miss the last letter.. – Weyland Yutani Feb 03 '14 at 13:46
Yeah, I mixed up my Random implementations, it's fixed now. – Winderps Feb 03 '14 at 13:51
no problem I had to remember it myself – Weyland Yutani Feb 03 '14 at 13:51

score 1 · Answer 3 · answered Feb 03 '14 at 13:45

How can you say rarely?

try this:

        String[] myArray = new String[] { "A", "B", "C" };

        Random rd = new Random();

        Int32 aCounter = 0;
        Int32 bCounter = 0;
        Int32 cCounter = 0;

        for (int i = 0; i < 25000; i++)
        {
            Int32 retVal = rd.Next(myArray.Length);
            switch (retVal)
            {
                case 0:
                    aCounter++;
                    break;
                case 1:
                    bCounter++;
                    break;
                case 2:
                    cCounter++;
                    break;
            }
        }

As expected in my test, as expected aCounter, bCounter and cCounter always have same values between 8000 and 8500.

Perahps you called your new Random().Next() too soon use a single object Random and call Next() each time on the same object

Thank you for the comment, I mainly want to know how random() itself works? Does it choose a number between 0 to 1 randomly and then multiply it by another value? — CowBoy, Feb 03 '14 at 13:55
Basically a Random it's an object built on a seed value. You can decide to call the constructor with your own seed or without. each times you call an overload of the Next() method it applys some calculation starting from the seed. Take a look at the official documentation http://msdn.microsoft.com/en-us/library/System.Random(v=vs.110).aspx — Francesco Milani, Feb 03 '14 at 14:06

How random function inside random class generates an integer?

3 Answers3