为什么这个随机的Next（）抛出异常？

Question

Hi I'm new to programming and am working from this book. This is an exercise at the end of one of the chapters. But my code throws this exception.

        Random rand = new Random();
        List<int> numbers = new List<int>();

        for (int i = 0; i < 1000; i++)
        {
            numbers[i] = rand.Next(1, 1001);
        }

        for (int i = 0; i < numbers.Count; i++)
        {
            listBox1.Items.Add(numbers[i]);
        }

Here is the error:

Answer 1

numbers is empty, so any index is invalid. Either use numbers.Add to add numbers, or better use a pre-sized array. Using a List is not necessary here. An array is just fine because the final size is known. It is also faster, although that does not seem to be an issue here.

Answer 2

ArgumentOutOfRangeException is only called on Random.Next when minValue is greater than maxValue. as according to: http://msdn.microsoft.com/en-us/library/2dx6wyd4.aspx#ddueExceptionsToggle

Your problem lays with numbers[i] as numbers has zero elements, meaning setting numbers[0] = random.Next(1, 1000); will return an ArgumentOutOfRangeException because there is no range yet. Use numbers.Add(random.Next(1, 1000)); instead

Just a hint, when working with indexers in array type objects (when you use object[key] ) and for loops you should always use the arrays length as the upper limit of the for loop, like so:

for(int i = 0; i < 1000 && i < numbers.Count(); i++){
    //Do stuff with i
}

In your case the loop will never happen as i is always going to be smaller then numbers.Count() unless you add items to it

Answer 3

Your numbers List dosent have any elements yet.

try this:

numbers.add(rand.Next(1, 1001));

Answer 4

Try

numbers.Add(rand.Next(1, 1001));

instead of

numbers[i] = rand.Next(1, 1001);

so you can have a dynamically re-sizing list.

Your list does not have anything at index position i so you receive ArgumentOutOfRangeException . You can only access an item in a collection via its index if there is something there at the given index.

Answer 5

Try with this 

List

   Random rand = new Random();
                List<int> numbers = new List<int>();
                //int [] nums ;
                for (int i = 0; i < 1000; i++)
                {
                    //nums[i] = rand.Next(0, 1000);
                    numbers.Add(rand.Next(0,1000));
                    //numbers[i] = rand.Next(1, 1001);
                }

------------------------OR ------------------------------------------

You can use Array also

Array

Random rand = new Random();
            List<int> numbers = new List<int>();
            int [] nums = new int [1000] ;
            for (int i = 0; i < 1000; i++)
            {
                nums[i] = rand.Next(0, 1000);                

            }