C 为什么OpenMP会加速单次迭代循环?
我使用的是来自的read基准,我只添加了两行:C 为什么OpenMP会加速单次迭代循环?,c,performance,gcc,memory,openmp,C,Performance,Gcc,Memory,Openmp,我使用的是来自的read基准,我只添加了两行: #pragma omp parallel for for(unsigned dummy = 0; dummy < 1; ++dummy) 在同一位置隐式声明具有相同的效果 完整代码: #include <stdio.h> #include <stdlib.h> #include <string.h> #include <time.h> unsigned long do_xor(const
#pragma omp parallel for
for(unsigned dummy = 0; dummy < 1; ++dummy)
在同一位置隐式声明具有相同的效果
完整代码:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
unsigned long do_xor(const unsigned long* p, unsigned long n)
{
unsigned long i, x = 0;
for(i = 0; i < n; ++i)
x ^= p[i];
return x;
}
int main()
{
unsigned long n, r, i;
unsigned long *p;
clock_t c0, c1;
double elapsed;
n = 1000 * 1000 * 1000; /* GB */
r = 100; /* repeat */
p = calloc(n/sizeof(unsigned long), sizeof(unsigned long));
c0 = clock();
#pragma omp parallel for
for(unsigned dummy = 0; dummy < 1; ++dummy)
for(i = 0; i < r; ++i) {
p[0] = do_xor(p, n / sizeof(unsigned long)); /* "use" the result */
printf("%4ld/%4ld\r", i, r);
fflush(stdout);
}
c1 = clock();
elapsed = (c1 - c0) / (double)CLOCKS_PER_SEC;
printf("Bandwidth = %6.3f GB/s (Giga = 10^9)\n", (double)n * r / elapsed / 1e9);
free(p);
}
时间报告的墙时间为3.4s vs 7.5s
GCC 7.3.0 Ubuntu性能差异的原因实际上不是代码上的差异,而是内存的映射方式。在快速情况下,您从零页读取,即所有虚拟地址都映射到单个物理页,因此无需从内存读取任何内容。在慢速情况下,它不会归零。有关详细信息,请参阅 另一方面,它不是由调用omp_get_num_线程或pragma itstelf引起的,而只是链接到OpenMP运行库。您可以根据需要使用-Wl,-no-fopenmp来确认这一点。如果只指定-fopenmp,但根本不使用它,链接器将忽略它
现在不幸的是,我仍然没有找到最后一个谜题:为什么链接到OpenMP会改变calloc关于零页面的行为。我可以重现。这不是优化,也不是时钟的问题。尽管如此,我建议您可以通过测量墙时间来改进问题,并添加实际测量结果以及您的CPU和内存规格。-fopenmp可能使用不同的线程安全版本的calloc?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
unsigned long do_xor(const unsigned long* p, unsigned long n)
{
unsigned long i, x = 0;
for(i = 0; i < n; ++i)
x ^= p[i];
return x;
}
int main()
{
unsigned long n, r, i;
unsigned long *p;
clock_t c0, c1;
double elapsed;
n = 1000 * 1000 * 1000; /* GB */
r = 100; /* repeat */
p = calloc(n/sizeof(unsigned long), sizeof(unsigned long));
c0 = clock();
#pragma omp parallel for
for(unsigned dummy = 0; dummy < 1; ++dummy)
for(i = 0; i < r; ++i) {
p[0] = do_xor(p, n / sizeof(unsigned long)); /* "use" the result */
printf("%4ld/%4ld\r", i, r);
fflush(stdout);
}
c1 = clock();
elapsed = (c1 - c0) / (double)CLOCKS_PER_SEC;
printf("Bandwidth = %6.3f GB/s (Giga = 10^9)\n", (double)n * r / elapsed / 1e9);
free(p);
}
gcc -O3 -Wall -fopenmp single_iteration.c && time taskset -c 0 ./a.out