Question

我有一台64位NUMA机器，有2个库。我知道如果1个银行的CPU试图访问其他银行的内存，会有性能下降。我想通过一些示例代码来测量它。我试着写下面的内容（CPU1和CPU7在不同的库中）

#define _GNU_SOURCE
#include <sched.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <string.h>

#define BUFSZ 1048576*500
#define SCRATCHSZ 1048576*10
int main()
{
  char *scratch;
  char *buf;
  char *buf2;
  struct timeval  tv1, tv2;
  cpu_set_t affinity_mask;

  CPU_ZERO(&affinity_mask);
  CPU_SET(1, &affinity_mask);
  if (sched_setaffinity(0, sizeof(cpu_set_t), &affinity_mask)) {
    return -1;
  }

  buf = malloc(BUFSZ);
  buf2 = malloc(BUFSZ);
  scratch = malloc(SCRATCHSZ);

  /* Let page fault happen */
  memset(buf, 0, BUFSZ);
  memset(buf2, 0, BUFSZ);
  /* clear the cpu cache */
  memset(scratch, 0, SCRATCHSZ);

  gettimeofday(&tv1, NULL);
  memcpy(buf2, buf, BUFSZ);
  gettimeofday(&tv2, NULL);
  printf ("Total time 1 = %f useconds\n",
           (double) (tv2.tv_usec - tv1.tv_usec)  +
           (double) (tv2.tv_sec - tv1.tv_sec)*1000000);

  free(buf2);
  free(scratch);

  CPU_ZERO(&affinity_mask);
  CPU_SET(7, &affinity_mask);
  if (sched_setaffinity(0, sizeof(cpu_set_t), &affinity_mask)) {
    return -1;
  }
  buf2 = malloc(BUFSZ);
  scratch = malloc(SCRATCHSZ);

  memset(buf2, 0, BUFSZ);
  memset(scratch, 0, SCRATCHSZ);

  gettimeofday(&tv1, NULL);
  memcpy(buf2, buf, BUFSZ);
  gettimeofday(&tv2, NULL);

  printf ("Total time 2 = %f useconds\n",
          (double) (tv2.tv_usec - tv1.tv_usec)  +
          (double) (tv2.tv_sec - tv1.tv_sec)*1000000);

  free(buf);
  free(buf2);
  free(scratch);
  return 0;
}

执行此操作时：

prmpt产品：＆gt; ./a.out总时间1 = 169583.000000 useconds总时间2 = 129527.000000 useconds

这不是我所期待的。时间2必须大于time1，就像time2 CPU7试图从不同的存储区访问存储器一样。有什么想法吗？

测量对NUMA机器的性能影响以进行银行间复制

0 个答案: