OS X文件名中的变音器（perl）(Umlauts in OS X file names (perl))

 尽管您尝试与屏障同步，但您在working_threads上确实存在竞争条件，这很容易导致迭代次数不等： 
 
thread 0                             | thread 1
...                                  | ...
while (working_threads > 0) [==true] | ...
    if (my_num == 1) [==true]        | ...
        working_threads = 0          | ...
                                     | while (working_threads > 0) [==false]
    [hangs waiting for barrier]      | [hangs trying to exit from parallel]
 
 要修复特定代码，您还必须在while-condition-check和working_threads = 0之间添加一个屏障。 
 
#pragma omp parallel
    {   
        int my_num=omp_get_thread_num()+1;
        int idle=false;
        while(working_threads>0) {
#pragma omp barrier      
            if(my_num==1)
                working_threads=0;
#pragma omp barrier                                                                                                                       
        }   
    }
 
 请注意，代码并不是最惯用或最优雅的解决方案。 根据您的具体工作共享问题，可能有更好的方法。 此外，您必须确保worker_threads仅由单个线程编写 - 或在写入时使用atomics。 

Despite your attempt to synchronize with the barrier, you do have a race condition on working_threads that can easily lead to unequal amount of iterations:
 
thread 0                             | thread 1
...                                  | ...
while (working_threads > 0) [==true] | ...
    if (my_num == 1) [==true]        | ...
        working_threads = 0          | ...
                                     | while (working_threads > 0) [==false]
    [hangs waiting for barrier]      | [hangs trying to exit from parallel]
 
To fix your specific code, you would have to also add a barrier between the while-condition-check and working_threads = 0.
 
#pragma omp parallel
    {   
        int my_num=omp_get_thread_num()+1;
        int idle=false;
        while(working_threads>0) {
#pragma omp barrier      
            if(my_num==1)
                working_threads=0;
#pragma omp barrier                                                                                                                       
        }   
    }
 
Note that the code is not exactly the most idiomatic or elegant solution. Depending on your specific work-sharing problem, there may be a better approach. Also you must ensure that worker_threads is written only by a single thread - or use atomics when writing.