C 错误输出中的简单 OpenMP For 循环

Question

试图让一个简单的 OpenMP 循环运行，但我总是得到奇怪的输出。它没有直接从 1 到 1000 列出，而是从 501 到 750，然后从 1 到 1000。我猜这是线程问题？我正在 VS2013 上编译和运行。

#include <stdio.h>
#include <math.h>

int main(void)
{
    int counter = 0;
    double root = 0;

    // OPEN MP SECTION
    printf("Open MP section: \n\n");
    getchar(); //Pause

#pragma omp parallel for 

    for (counter = 0; counter <= 1000; counter++)
    {
        root = sqrt(counter);
        printf("The root of %d is %.2f\n", counter, root);

    }


    return(0);


}

Answer 1

将结果存储在一个数组中，并使 printf 退出循环。它必须序列化到显示器。

Answer 2

OpenMP 的全部要点是运行并行，将工作分配给不同的执行引擎。

因此，您的循环的各个迭代很可能是乱序完成的，因为这是多线程的本质。

虽然计算并行完成可能有意义（因此可能会乱序），但这并不是您真正想要的打印个结果。

确保结果按正确顺序打印的一种方法是将打印推迟到并行执行完成之后。换句话说，并行化计算但串行化输出。

这当然意味着能够将信息存储在例如数组中，而并行操作是运行ning。

换句话说，类似于：

// Make array instead of single value.

double root[1001];

// Parallelise just the calculation bit.

#pragma omp parallel for 
for (counter = 0; counter <= 1000; counter++)
    root[counter] = sqrt(counter);

// Leave the output as a serial operation,
//   once all parallel operations are done.

for (counter = 0; counter <= 1000; counter++)
    printf("The root of %d is %.2f\n", counter, root[counter]);

Answer 3

您的代码不会运行顺序。

OpenMP 并行编译指示：

#pragma omp parallel
{
  // Code inside this region runs in parallel.
  printf("Hello!\n");
}

'This code creates a team of threads, and each thread executes the same code. It prints the text "Hello!" followed by a newline, as many times as there are threads in the team created. For a dual-core system, it will output the text twice. (Note: It may also output something like "HeHlellolo", depending on system, because the printing happens in parallel.) At the }, the threads are joined back into one, as if in non-threaded program."'

http://bisqwit.iki.fi/story/howto/openmp/#ParallelPragma

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