C# Parallel.For unexpected results

Question

I am a newbie in C# and while translating my old DSP, VB6 code I would like to take advantage of the possibilities offered by this language. Specifically I am trying to use parallel processing to speed up time expensive calculations.

Here is the (simplified) code that I am currently testing.

The "1° parallel code OK" section works fine, no problems;

the "2° parallel code WRONG results" section that, to me, should be equivalent to the "Sequential code OK" section is a lot faster than the sequential version but doesn't return the expected results.

Why? Where am I wrong?

     double[] Omega = new double[NFreq + 1];
     double[,] SinOmT = new double[NFreq + 1, NVAL];
     double[,] CosOmT = new double[NFreq + 1, NVAL];

 double OmT;

    // 1° Parallel code OK:
      Parallel.For(0, NFreq + 1, JF =>
    {
        Omega[JF] = PI2 * Freq[JF] / Fs;
    });

    // Sequential code OK:
    for (int JF = 1; JF <= NFreq; JF++)
    {
        for (int JD = 0; JD < NVAL; JD++)
        {
            OmT = Omega[JF] * (double)(JD);
            SinOmT[JF, JD] = Math.Sin(OmT);
            CosOmT[JF, JD] = Math.Cos(OmT);
        }
    }

    // 2° Parallel code WRONG results:
    for (int JF = 1; JF <= NFreq; JF++)
    {
        Parallel.For(0, NVAL, JD =>
        {
            OmT = Omega[JF] * (double)(JD);
            SinOmT[JF, JD] = Math.Sin(OmT);
            CosOmT[JF, JD] = Math.Cos(OmT);
        });
    }

Ciao and thanks for the attention.

Franco

score 4 · Answer 1 · answered Jun 07 '14 at 20:45

4

It's wrong to declare the variable OmT outside of the loops, because then the parallel iterations contest for it and overwrite its value unpredictably.

In C#, you should always declare variables in the tightest scope that makes sense. This is true for all scenarios, not just parallel code.

        var OmT = Omega[JF] * (double)(JD);
        SinOmT[JF, JD] = Math.Sin(OmT);
        CosOmT[JF, JD] = Math.Cos(OmT);

answered Jun 07 '14 at 20:45

Theodoros Chatzigiannakis

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Theodoros: it works GREAT !! Thank you so much. Ciao and to the next. – Franco Languasco Jun 08 '14 at 11:11

score 1 · Answer 2 · edited May 23 '17 at 12:11

1

The variable JF gets captured by the lambda expression; this means that it will always have the same value. See here for a better explaination.

edited May 23 '17 at 12:11

Community

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answered Jun 07 '14 at 20:44

BlackBear

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C# Parallel.For unexpected results

2 Answers2