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linux - Pthreads 程序比串行程序慢 - Linux

问题描述

感谢您在这件事上慷慨解囊并帮助我。我正在尝试使用 pthread 计算平方数的总和。但是，它似乎比串行实现还要慢。此外，当我增加线程数时，程序变得更慢。我确保每个线程都在不同的核心上运行（我有 6 个核心分配给虚拟机）

这是串行程序：

#include <stdio.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/time.h>
#include <time.h>


int main(int argc, char *argv[]) {
  
  struct timeval start, end;
  gettimeofday(&start, NULL);   //start time of calculation
  
   int n = atoi(argv[1]);
   long int sum = 0;
   for (int i = 1; i < n; i++){
      sum += (i * i);
   }
  
   gettimeofday(&end, NULL); //end time of calculation
   printf("The sum of squares in [1,%d): %ld | Time Taken: %ld mirco seconds \n",n,sum, 
   ((end.tv_sec * 1000000 + end.tv_usec) - (start.tv_sec * 1000000 + start.tv_usec)));

   return 0;
}

这是 Pthreads 程序：

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <sys/types.h>
#include <sys/time.h>
#include <time.h>


void *Sum(void *param);

// structure for thread arguments
struct thread_args {
int tid;
int a; //start
int b; //end
long int result; // partial results
};


int main(int argc, char *argv[])
{
struct timeval start, end;
gettimeofday(&start, NULL);   //start time of calculation

int numthreads;
int number;
double totalSum=0;

if(argc < 3 ){
printf("Usage: ./sum_pthreads <numthreads> <number> ");
return 1;
}

numthreads = atoi(argv[1]);
number = atoi(argv[2]);;
pthread_t   tid[numthreads];
struct thread_args targs[numthreads];

printf("I am Process | range: [%d,%d)\n",1,number);
printf("Running Threads...\n\n");

for(int i=0; i<numthreads;i++ ){
//Setting up the args
targs[i].tid = i;
targs[i].a = (number)*(targs[i].tid)/(numthreads);
targs[i].b = (number)*(targs[i].tid+1)/(numthreads);
if(i == numthreads-1 ){
targs[i].b = number;
}

pthread_create(&tid[i],NULL,Sum, &targs[i]);

}


for(int i=0; i< numthreads; i++){
pthread_join(tid[i],NULL);
}

printf("Threads Exited!\n");
printf("Process collecting information...\n");

for(int i=0; i<numthreads;i++ ){
  totalSum += targs[i].result;
}

gettimeofday(&end, NULL); //end time of calculation

 
printf("Total Sum is: %.2f | Taken Time: %ld mirco seconds \n",totalSum, 
((end.tv_sec * 1000000 + end.tv_usec) - (start.tv_sec * 1000000 + start.tv_usec)));

return 0;
}



void *Sum(void *param) {
int start = (*(( struct  thread_args*) param)).a;
int end = (*((struct  thread_args*) param)).b;
int id = (*((struct thread_args*)param)).tid;
long int sum =0;

printf("I am thread %d | range: [%d,%d)\n",id,start,end);

for (int i = start; i < end; i++){
      sum += (i * i);
   }
   
(*((struct thread_args*)param)).result =  sum;
printf("I am thread %d | Sum: %ld\n\n", id ,(*((struct thread_args*)param)).result );
pthread_exit(0);
}

结果：

hamza@hamza:~/Desktop/lab4$ ./sum_serial 10
The sum of squares in [1,10): 285 | Time Taken: 7 mirco seconds 
hamza@hamza:~/Desktop/lab4$ ./sol 2 10
I am Process | range: [1,10)
Running Threads...

I am thread 0 | range: [0,5)
I am thread 0 | Sum: 30

I am thread 1 | range: [5,10)
I am thread 1 | Sum: 255

Threads Exited!
Process collecting information...
Total Sum is: 285.00 | Taken Time: 670 mirco seconds 
hamza@hamza:~/Desktop/lab4$ ./sol 3 10
I am Process | range: [1,10)
Running Threads...

I am thread 0 | range: [0,3)
I am thread 0 | Sum: 5

I am thread 1 | range: [3,6)
I am thread 1 | Sum: 50

I am thread 2 | range: [6,10)
I am thread 2 | Sum: 230

Threads Exited!
Process collecting information...
Total Sum is: 285.00 | Taken Time: 775 mirco seconds 
hamza@hamza:~/Desktop/lab4$

标签： linuxmultithreadingubuntuparallel-processingpthreads

解决方案

这两个程序做的事情非常不同。例如，线程程序会产生更多的文本输出并创建一堆线程。您正在比较非常短的运行（不到千分之一秒），因此这些额外事情的开销很大。

您必须以更长的运行时间进行测试，这样才能产生额外的输出以及创建和同步线程的成本。

打个比方，一个人可以比三个人更快地拧紧三个螺钉，因为每个人都需要一个工具，决定谁来拧紧哪个螺钉等等。但是如果你有 500 颗螺丝要拧紧，那么三个人会更快地完成它。

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