c - 为什么我的 OpenMP 代码性能比串行的差？

Question

我正在做一个简单的 Pi 计算，我将生成随机数并递增计数的循环并行化。串行(非 OpenMP)代码的性能优于 OpenMP 代码。这是我进行的一些测量。下面还提供了这两个代码。

编译串口代码为:gcc pi.c -O3

将 OpenMP 代码编译为:gcc pi_omp.c -O3 -fopenmp

可能是什么问题？

# Iterations = 60000000

Serial Time = 0.893912

OpenMP 1 Threads Time = 0.876654
OpenMP 2 Threads Time = 23.8537
OpenMP 4 Threads Time = 7.72415

序列号:

/* Program to compute Pi using Monte Carlo methods */
/* from: http://www.dartmouth.edu/~rc/classes/soft_dev/C_simple_ex.html */

#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <time.h>
#include <sys/time.h>
#define SEED 35791246

int main(int argc, char* argv)
{
  int niter=0;
  double x,y;
  int i;
  long count=0; /* # of points in the 1st quadrant of unit circle */
  double z;
  double pi;

  printf("Enter the number of iterations used to estimate pi: ");
  scanf("%d",&niter);

  /* initialize random numbers */
  srand(SEED);
  count=0;
  struct timeval start, end;
  gettimeofday(&start, NULL);
  for ( i=0; i<niter; i++) {
    x = (double)rand()/RAND_MAX;
    y = (double)rand()/RAND_MAX;
    z = x*x+y*y;
    if (z<=1) count++;
  }
  pi=(double)count/niter*4;

  gettimeofday(&end, NULL);
  double t2 = end.tv_sec + (end.tv_usec/1000000.0);
  double t1 = start.tv_sec + (start.tv_usec/1000000.0);

  printf("Time: %lg\n", t2 - t1);

  printf("# of trials= %d , estimate of pi is %lg \n",niter,pi);
  return 0;
}

OpenMP 并行代码:

#include <omp.h>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <time.h>
#include <sys/time.h>
#define SEED 35791246
/*
from: http://www.dartmouth.edu/~rc/classes/soft_dev/C_simple_ex.html
 */
#define CHUNKSIZE 500
int main(int argc, char *argv[]) {

  int chunk = CHUNKSIZE;
  int niter=0;
  double x,y;
  int i;
  long count=0; /* # of points in the 1st quadrant of unit circle */
  double z;
  double pi;

  int nthreads, tid;

  printf("Enter the number of iterations used to estimate pi: ");
  scanf("%d",&niter);

  /* initialize random numbers */
  srand(SEED);
  struct timeval start, end;

  gettimeofday(&start, NULL);
  #pragma omp parallel shared(chunk) private(tid,i,x,y,z) reduction(+:count)  
  {                                                                                                           
    /* Obtain and print thread id */
    tid = omp_get_thread_num();
    //printf("Hello World from thread = %d\n", tid);

    /* Only master thread does this */
    if (tid == 0)
    {
      nthreads = omp_get_num_threads();
      printf("Number of threads = %d\n", nthreads);
    }

    #pragma omp for schedule(dynamic,chunk)                                                                       
    for ( i=0; i<niter; i++) {                                                                              
      x = (double)rand()/RAND_MAX;                                                                          
      y = (double)rand()/RAND_MAX;                                                                          
      z = x*x+y*y;                                                                                          
      if (z<=1) count++;                                                                                    
    }                                                                                                       
  }                                                                                                           

  gettimeofday(&end, NULL);
  double t2 = end.tv_sec + (end.tv_usec/1000000.0);
  double t1 = start.tv_sec + (start.tv_usec/1000000.0);

  printf("Time: %lg\n", t2 - t1);

  pi=(double)count/niter*4;                                                                                   
  printf("# of trials= %d, threads used: %d, estimate of pi is %lg \n",niter,nthreads, pi);
  return 0;
}

Answer 1

rand() 不可重入。它要么无法正常工作、崩溃，要么一次只能从一个线程调用。像 glibc 这样的库通常会为遗留的不可重入函数序列化或使用 TLS，而不是让它们在多线程代码中使用时随机崩溃。

尝试重入形式，rand_r():

tid = omp_get_thread_num();
unsigned int seed = tid;
...
x = (double)rand_r(&seed)/RAND_MAX;

我想您会发现它要快得多。

请注意我是如何将种子设置为 tid 的。您可能会想，为什么不将种子初始化为 SEED？给定相同的种子，rand_r() 将产生相同的数字序列。如果每个线程都使用同一系列的伪随机数，那么它就失去了进行更多迭代的意义!您必须让每个线程使用不同的数字。