Average function without overflow exception

Posted by Ron Klein on Stack Overflow See other posts from Stack Overflow or by Ron Klein
Published on 2010-05-24T07:58:25Z Indexed on 2010/05/24 13:41 UTC
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.NET Framework 3.5.
I'm trying to calculate the average of some pretty large numbers.
For instance:

using System;
using System.Linq;

class Program
{
    static void Main(string[] args)
    {
        var items = new long[]
                        {
                            long.MinValue + 100, 
                            long.MinValue + 200, 
                            long.MinValue + 300
                        };
        try
        {
            var avg = items.Average();
            Console.WriteLine(avg);
        }
        catch (OverflowException ex)
        {
            Console.WriteLine("can't calculate that!");
        }
        Console.ReadLine();
    }
}

Obviously, the mathematical result is 9223372036854775607 (long.MaxValue - 200), but I get an exception there. This is because the implementation (on my machine) to the Average extension method, as inspected by .NET Reflector is:

public static double Average(this IEnumerable<long> source)
{
    if (source == null)
    {
        throw Error.ArgumentNull("source");
    }
    long num = 0L;
    long num2 = 0L;
    foreach (long num3 in source)
    {
        num += num3;
        num2 += 1L;
    }
    if (num2 <= 0L)
    {
        throw Error.NoElements();
    }
    return (((double) num) / ((double) num2));
}

I know I can use a BigInt library (yes, I know that it is included in .NET Framework 4.0, but I'm tied to 3.5).

But I still wonder if there's a pretty straight forward implementation of calculating the average of integers without an external library. Do you happen to know about such implementation?

Thanks!!


UPDATE:

The previous example, of three large integers, was just an example to illustrate the overflow issue. The question is about calculating an average of any set of numbers which might sum to a large number that exceeds the type's max value. Sorry about this confusion. I also changed the question's title to avoid additional confusion.

Thanks all!!

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