C++ OpenMP并行化中的分段故障

我正在使用OpenMP并行以下函数

float myfunc ( Class1 *class1, int *feas, int numfeas, float z, long *k, double cost, long iter, float e )
{
    long i;
    long x;
    double sum;

    sum = cost;
    while ( change/cost > 1.0*e ) {
        change = 0.0;

        intshuffle ( feas, numfeas );

        #pragma omp parallel for private (i,x) firstprivate(z,k) reduction (+:sum)
        for ( i=0; i<iter; i++ ) {
            x = i%numfeas;
            sum += pgain ( feas[x], class1, z, k );
        }
        cost -= sum;
    }
    return ( cost );
}

我得到一个“双重免费或损坏（输出）：0xb6a00468”错误

根据我对OpenMp的有限知识，我理解这是由于对class1和feas指针的错误内存访问造成的

作为参考，我还张贴了我的Class1代码

typedef struct {
    float w;
    float *c;
    long a;  
    float co;  
} Class1;

请就并联上述功能的正确方式提出建议

更新：

pgain函数

double pgain ( long x, Class1 *class1, double z, long int *numcenters )
{
    int i;
    int number_of_centers_to_close = 0;

    static double *work_mem;
    static double gl_cost_of_opening_x;
    static int gl_number_of_centers_to_close;

    int stride = *numcenters + 2;
    //make stride a multiple of CACHE_LINE
    int cl = CACHE_LINE/sizeof ( double );
    if ( stride % cl != 0 ) {
        stride = cl * ( stride / cl + 1 );
    }
    int K = stride - 2 ; // K==*numcenters

    //my own cost of opening x
    double cost_of_opening_x = 0;

    work_mem = ( double* ) malloc ( 2 * stride * sizeof ( double ) );
    gl_cost_of_opening_x = 0;
    gl_number_of_centers_to_close = 0;

    /*
     * For each center, we have a *lower* field that indicates
     * how much we will save by closing the center.
     */
    int count = 0;
    for ( int i = 0; i < class1->num; i++ ) {
        if ( is_center[i] ) {
            center_table[i] = count++;
        }
    }
    work_mem[0] = 0;

    //now we finish building the table. clear the working memory.
    memset ( switch_membership, 0, class1->num * sizeof ( bool ) );
    memset ( work_mem, 0, stride*sizeof ( double ) );
    memset ( work_mem+stride,0,stride*sizeof ( double ) );

    //my *lower* fields
    double* lower = &work_mem[0];
    //global *lower* fields
    double* gl_lower = &work_mem[stride];

    #pragma omp parallel for 
    for ( i = 0; i < class1->num; i++ ) {
        float x_cost = dist ( class1->p[i], class1->p[x], class1->dim ) * class1->p[i].weight;
        float current_cost = class1->p[i].cost;

        if ( x_cost < current_cost ) {

            // point i would save cost just by switching to x
            // (note that i cannot be a median,
            // or else dist(p[i], p[x]) would be 0)

            switch_membership[i] = 1;
            cost_of_opening_x += x_cost - current_cost;

        } else {

            // cost of assigning i to x is at least current assignment cost of i

            // consider the savings that i's **current** median would realize
            // if we reassigned that median and all its members to x;
            // note we've already accounted for the fact that the median
            // would save z by closing; now we have to subtract from the savings
            // the extra cost of reassigning that median and its members
            int assign = class1->p[i].assign;
            lower[center_table[assign]] += current_cost - x_cost;
        }
    }

    // at this time, we can calculate the cost of opening a center
    // at x; if it is negative, we'll go through with opening it

    for ( int i = 0; i < class1->num; i++ ) {
        if ( is_center[i] ) {
            double low = z + work_mem[center_table[i]];
            gl_lower[center_table[i]] = low;
            if ( low > 0 ) {
                // i is a median, and
                // if we were to open x (which we still may not) we'd close i

                // note, we'll ignore the following quantity unless we do open x
                ++number_of_centers_to_close;
                cost_of_opening_x -= low;
            }
        }
    }
    //use the rest of working memory to store the following
    work_mem[K] = number_of_centers_to_close;
    work_mem[K+1] = cost_of_opening_x;

    gl_number_of_centers_to_close = ( int ) work_mem[K];
    gl_cost_of_opening_x = z + work_mem[K+1];

    // Now, check whether opening x would save cost; if so, do it, and
    // otherwise do nothing

    if ( gl_cost_of_opening_x < 0 ) {
        //  we'd save money by opening x; we'll do it
        #pragma omp parallel for
        for ( int i = 0; i < class1->num; i++ ) {
            bool close_center = gl_lower[center_table[class1->p[i].assign]] > 0 ;
            if ( switch_membership[i] || close_center ) {
                // Either i's median (which may be i itself) is closing,
                // or i is closer to x than to its current median
                #pragma omp critical
                {
                class1->p[i].cost = class1->p[i].weight * dist ( class1->p[i], class1->p[x], class1->dim );
                class1->p[i].assign = x;
                }
            }
        }
        for ( int i = 0; i < class1->num; i++ ) {
            if ( is_center[i] && gl_lower[center_table[i]] > 0 ) {
                is_center[i] = false;
            }
        }
        if ( x >= 0 && x < class1->num ) {
            is_center[x] = true;
        }

        *numcenters = *numcenters + 1 - gl_number_of_centers_to_close;
    } else {
        gl_cost_of_opening_x = 0;  // the value we'll return
    }

    free ( work_mem );

    return -gl_cost_of_opening_x;
}

---

问题可能是

pgain

的这一部分中可能存在竞态：

// Either i's median (which may be i itself) is closing,
// or i is closer to x than to its current median
class1->p[i].cost = class1->p[i].weight * dist ( class1->p[i], class1->p[x], class1->dim );
class1->p[i].assign = x;

由于

class1

是一个指针，因此使用语句

firstprivate（class1）

将私有化的是裸指针，而不是底层资源

这个问题的解决方案在很大程度上取决于程序的语义。如果可以对

*class1

进行随机排序的更新，并且该资源确实是要共享的，则需要进行以下修改：

// Either i's median (which may be i itself) is closing,
// or i is closer to x than to its current median
#pragma omp critical LOCK_CLASS1
{
  // Lock this part of the code for thread-safety
  class1->p[i].cost = class1->p[i].weight * dist ( class1->p[i], class1->p[x], class1->dim );
  class1->p[i].assign = x
}

将修复上面的代码。否则，在调用

pgain

之前，应该为每个线程创建

*class1

的私有副本

---

最后，您应该注意，上述推理对于任何资源都是正确的。例如，

Class1

中的指针

float*c

显示相同的临界点，如果它指向共享的资源，并且您不同步内存更新。

您确定

i

必须是私有的吗？我很确定您可以省去该指令，而内联

x

。我需要回顾一下我的OpenMP笔记…@Mysticial pgain是另一个函数签名是double pgain（长x，Class1*Class1，双z，长int*numcent）。但这并不重要，因为代码在没有OpenMP的情况下可以正确编译和运行pragmas@millinon我不需要是私有的，因为OpenMp足够聪明，可以识别它是私有的，但是x需要是私有的。问题可能在于共享内存以及feas和class1的端口。您是否有可能在该函数中分配内存？内存分配器是线程安全的吗？@Mysticial是的，我在PGAIN中运行malloc。那么您建议的parellization问题解决方案是什么？无法修复它。我已经用新信息更新了我原来的帖子。我接受你的回复。赛格出现故障的原因确实是赛道状况。

// Either i's median (which may be i itself) is closing,
// or i is closer to x than to its current median
class1->p[i].cost = class1->p[i].weight * dist ( class1->p[i], class1->p[x], class1->dim );
class1->p[i].assign = x;

// Either i's median (which may be i itself) is closing,
// or i is closer to x than to its current median
#pragma omp critical LOCK_CLASS1
{
  // Lock this part of the code for thread-safety
  class1->p[i].cost = class1->p[i].weight * dist ( class1->p[i], class1->p[x], class1->dim );
  class1->p[i].assign = x
}