我的代码旨在将两个数组组合成第三个数组，并多次执行此操作。我希望使用OpenMP的方式将其并行化，但结果并不是我所期望的。也就是说，第一个检查已经在a[0]！=0处中断。
MWE是：

#include <stdio.h>
#include <stdlib.h>
#include <omp.h>

int main(int argc, char *argv[])
{
    int i = 0;
    int j = 0;
    long length = 10000;
    long nSamples = 1000;
    double *a, *b, *c;
    double start = 0;
    double end = 0;
    int nthreads = 0;

// Array Allocation and Initialisation
    a = (double *)malloc(length * sizeof(double));
    b = (double *)malloc(length * sizeof(double));
    c = (double *)malloc(length * sizeof(double));
    for (i = 0; i < length; i++)
    {
        a[i] = 0.0;
        b[i] = 0.0;
        c[i] = 0.0;
    }

//  Get the maximum number of threads and evaluate the number of threads per group
#pragma omp parallel shared(nthreads)
    {
        nthreads = omp_get_num_threads() / 2;
    }

#pragma omp parallel for shared(nthreads, a, b, c) num_threads(nthreads) collapse(2)
    for (i = 0; i < nSamples; i++)
    {
        for (j = 0; j < length; j++)
        {
            a[j] = a[j] + 2.0;
            b[j] = b[j] + 2.0;
            c[j] = a[j] + b[j];
        }
    }

    /*Check correctness*/
    for (i = 0; i < length; i++)
    {
        if (a[i] != 2.0 * nSamples)
        {
            printf("a not equal at %d\n", i);
            break;
        }
        if (b[i] != 2.0 * nSamples)
        {
            printf("b ot equal at %d\n", i);
            break;
        }
        if (c[i] != 4.0 * nSamples)
        {
            printf("c ot equal at %d\n", i);
            break;
        }
    }

    free(a);
    free(b);
    free(c);
    return 0;
}

如果我只并行化内部循环，检查就能正常工作。为了实现这一点，我用途：

for (i = 0; i < nSamples; i++)
    {
#pragma omp parallel for private(j) shared(a, b, c) num_threads(nthreads)
        for (j = 0; j < length; j++)
        {
            a[j] = a[j] + 2.0;
            b[j] = b[j] + 2.0;
            c[j] = a[j] + b[j];
        }
    }

我知道openMP将循环的每次迭代视为独立于其他迭代。我假设这会导致线程1上的迭代i=5写入另一个线程的迭代i=9的a[j]，从而导致差异。这是正确的吗？使用openMP的方法可以避免这种情况吗？