我不认为你这样做 - 我为解决这个问题所做的是使用并行区域#pragma omp parallel shared(...) private(...)
并在并行区域内动态分配数组。试试这个:
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
/* compile with gcc -o test2 -fopenmp test2.c */
int main(int argc, char** argv)
{
int i = 0;
int size = 20;
int* a = (int*) calloc(size, sizeof(int));
int* b = (int*) calloc(size, sizeof(int));
int* c;
for ( i = 0; i < size; i++ )
{
a[i] = i;
b[i] = size-i;
printf("[BEFORE] At %d: a=%d, b=%d\n", i, a[i], b[i]);
}
#pragma omp parallel shared(a,b) private(c,i)
{
c = (int*) calloc(3, sizeof(int));
#pragma omp for
for ( i = 0; i < size; i++ )
{
c[0] = 5*a[i];
c[1] = 2*b[i];
c[2] = -2*i;
a[i] = c[0]+c[1]+c[2];
c[0] = 4*a[i];
c[1] = -1*b[i];
c[2] = i;
b[i] = c[0]+c[1]+c[2];
}
free(c);
}
for ( i = 0; i < size; i++ )
{
printf("[AFTER] At %d: a=%d, b=%d\n", i, a[i], b[i]);
}
}
这对我来说产生了与我之前的实验程序相同的结果:
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
/* compile with gcc -o test1 -fopenmp test1.c */
int main(int argc, char** argv)
{
int i = 0;
int size = 20;
int* a = (int*) calloc(size, sizeof(int));
int* b = (int*) calloc(size, sizeof(int));
for ( i = 0; i < size; i++ )
{
a[i] = i;
b[i] = size-i;
printf("[BEFORE] At %d: a=%d, b=%d\n", i, a[i], b[i]);
}
#pragma omp parallel for shared(a,b) private(i)
for ( i = 0; i < size; i++ )
{
a[i] = 5*a[i]+2*b[i]-2*i;
b[i] = 4*a[i]-b[i]+i;
}
for ( i = 0; i < size; i++ )
{
printf("[AFTER] At %d: a=%d, b=%d\n", i, a[i], b[i]);
}
}
我猜测是因为 OpenMP 无法推断出数组的大小,所以它不能是私有的 - 只有编译时数组才能以这种方式完成。当我尝试私有一个动态分配的数组时,我得到了段错误,大概是因为访问冲突。在每个线程上分配数组,就像您使用 pthreads 编写的一样有意义并解决了问题。