Random numbers in C#

Question

I get the error below when I try to run the application I am sure its something simple but I dont see it. What I am trying to do it when I click a button I have labeled Play. I

Answer 1

Setup one Random object and initialize it once.

class Form1
{
    ...
    Random rnd = new Random();
}

then to use it every time it is needed

void RandomNumber(int min, int max)
{
    int num = rnd.Next(min, max);
    ...
}

What happens everytime you call new() it re-seeds the random number and you may end up with the same numbers over and over. I have had this happend to me and it killed me

Answer 2

Well, your code doesn't match the error, but look at this:

private void btnPlay_Click(object sender, EventArgs e)
{
    RandomNumber();
}

private void RandomNumber(int min, int max)
{
    int num = new Random().Next(min, max);
    lblPickFive_1.Text = num.ToString();
}

RandomNumber has two parameters, min and max. You haven't specified any in the call inside btnPlay_Click. That's what the compiler is complaining about. Change the call to something like:

RandomNumber(5, 10);

Even when that's fixed, you shouldn't create a new instance of Random each time. As it happens, it's unlikely to cause problems in this particular case as it's triggered by a user action, but you should read my article on random numbers to see what the problem is and how to avoid it.

Answer 3

You need to pass in values:

private void btnPlay_Click(object sender, EventArgs e)
{
    RandomNumber();

}

should be:

private void btnPlay_Click(object sender, EventArgs e)
{
    RandomNumber(0, 50000);

}

Answer 4

First, you shouldn't be newing up a random number generator every time you want a new random number. You should set the generator as a static or member variable and refer to it for each new number.

Second, you have to pass a min and a max to your RandomNumber method.

Answer 5

You're calling RandomNumber(); in btnPlay_Click but the RandomNumber method requires min and max.

Answer 6

Your RandomNumber method takes two arguments.

If you want to call the method, you need to pass two arguments.