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Using Random

Started by foolios Aug 5, 2007 at 4:39 PM 4 replies 1.7k views

Original Post

Aug 5, 2007 at 4:39 PM

I want to create a random number based on a range, but for some reason it isn't working. http://msdn2.microsoft.com/en-us/library/2dx6wyd4.aspx I tried to figure this out from some examples in the MSDN. It appears there is a NEXT method that can be used with random. The intellisense tells me that I can use a minimum range and a maximum range. Take a look please:


            int r = h * w;
            Room[] rooms = new Room[r];
            Random rm = new Random();
            rm.Next(0, r);
            Console.WriteLine(rm);

Instead of the random number, the text System.Random is what's printed to the screen.

foolios

Author

foolios

Aug 5, 2007 at 4:45 PM

Ok, somehow I realized that I needed to add the line:

int newrm = rm.Next(0, r);

so that the newrm now has this new random number.
But I don't understand why I had to do this. rm was already an object. Why did I need to assign rm.Next() to a variable?
Thanks in advance.

Telastyn

268

Telastyn

Aug 5, 2007 at 4:46 PM

Random.Next(...) returns the random number. The class Random represents a random number generator, not a random number itself.

foolios

Author

foolios

Aug 5, 2007 at 4:48 PM

Ok, after looking it over a couple time, I see that .Next is a method of the random class and in that code, the method is being run from the new rm object. It has a return value and I must assign it to something if I want to use it.

foolios

Author

foolios

Aug 5, 2007 at 4:49 PM

Quote:
Original post by Telastyn
Random.Next(...) returns the random number. The class Random represents a random number generator, not a random number itself.

Thank you

borto

Aug 5, 2007 at 5:10 PM

You also don't need to set the minimum range if it's going to be 0. Random.Next with one parameter will return a range from zero to one below that number.

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