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Home/ Questions/Q 201829
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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T17:14:01+00:00 2026-05-11T17:14:01+00:00

Can anyone tell me what the correct Plinq code is for this? I’m adding

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Can anyone tell me what the correct Plinq code is for this? I’m adding up the square root of the absolute value of the sine of each element fo a double array, but the Plinq is giving me the wrong result.

Output from this program is:

Linq aggregate = 75.8310477905274 (correct)
Plinq aggregate = 38.0263653589291 (about half what it should be)

I must be doing something wrong, but I can’t work out what…

(I’m running this with Visual Studio 2008 on a Core 2 Duo Windows 7 x64 PC.)

Here’s the code:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Collections;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main()
        {
            double[] array = new double[100];

            for (int i = 0; i < array.Length; ++i)
            {
                array[i] = i;
            }

            double sum1 = array.Aggregate((total, current) => total + Math.Sqrt(Math.Abs(Math.Sin(current))));
            Console.WriteLine("Linq aggregate = " + sum1);

            IParallelEnumerable<double> parray = array.AsParallel<double>();
            double sum2 = parray.Aggregate((total, current) => total + Math.Sqrt(Math.Abs(Math.Sin(current))));
            Console.WriteLine("Plinq aggregate = " + sum2);
        }
    }
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-11T17:14:01+00:00Added an answer on May 11, 2026 at 5:14 pm

    Aggregate works slightly differently in PLINQ.

    From MSDN Blogs:

    Rather than expecting a value to
    initialize the accumulator to, the
    user gives us a factory function that
    generates the value:

    public static double Average(this IEnumerable<int> source)
    {
        return source.AsParallel().Aggregate(
            () => new double[2],
            (acc, elem) => { acc[0] += elem; acc[1]++; return acc; },
            (acc1, acc2) => { acc1[0] += acc2[0]; acc1[1] += acc2[1]; return acc1; },
            acc => acc[0] / acc[1]);
    }
    

    Now, PLINQ can initialize an
    independent accumulator for each
    thread. Now that each thread gets its
    own accumulator, both the folding
    function and the accumulator combining
    function are free to mutate the
    accumulators. PLINQ guarantees that
    accumulators will not be accessed
    concurrently from multiple threads .

    So, in your case, you would also need to pass an accumulator function which sums the outputs of the paralleled aggregates (hence why you’re seeing a result that is roughly half of what it should be).

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