我使用OpenMP来并行以下函数
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 );
}
这将返回“分段错误（核心转储）”错误
当我将我的编译指示更改为
时#pragma omp parallel for private (i,x) firstprivate(class1,feas,z,k) reduction (+:sum)`
我收到“双重免费或损坏（退出）：0xb6a00468 ”错误。
根据我对OpenMp的有限知识，我明白这是由于对 class1 和可行性指针的错误内存访问。
作为参考，我也发布了我的Class1代码
typedef struct {
float w;
float *c;
long a;
float co;
} Class1;
请告知有关并行化上述功能的正确方法。
更新
pgain功能
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;
}
更新2
我的版本修复了pgain原因我在pgain中也有#pragma指令
这是一个堆栈跟踪
#0 0xb7d50750 in ?? () from /lib/i386-linux-gnu/libc.so.6
#1 0xb7eaa198 in ?? () from /usr/lib/i386-linux-gnu/libgomp.so.1
#2 0x080493a0 in pgain (x=2982, class1=0xbffff218, z=1461.919921875,
numce=0x804d128) at streamcluster-openmp.cpp:232
#3 0x0804aaf6 in myfunc(Class1*, int*, int, float, long*, double, long, float) [clone ._omp_fn.0] () at openmp.cpp:347
#4 0xb7ea9889 in ?? () from /usr/lib/i386-linux-gnu/libgomp.so.1
#5 0xb7cbcd4c in start_thread () from /lib/i386-linux-gnu/libpthread.so.0
#6 0xb7dc9bae in clone () from /lib/i386-linux-gnu/libc.so.6
答案 0 :(得分：2)
问题可能是pgain的这一部分可能存在的竞争条件：
// 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，你使用语句firstprivate(class1)私有的是裸指针，而不是底层资源。
此问题的解决方案在很大程度上取决于程序的语义。如果可以随机排序*class1的更新，并且资源真的要共享，那么进行以下修改：
// 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，您应该在调用*class1之前为每个线程创建pgain的私有副本。
最后，您应该注意，对于任何资源，上述推理仍然适用。例如，float *c中的指针Class1如果引用共享的资源并且您不同步内存更新，则显示相同的临界点。