问题

大家好，我有一个程序（来自网络），我打算通过使用将其转换为并行版本来加快速度，pthreads但它的运行速度比串行版本慢。下面是程序：

# include <stdio.h>

//fast square root algorithm
double asmSqrt(double x) 
{
  __asm__ ("fsqrt" : "+t" (x));
  return x;
}

//test if a number is prime
bool isPrime(int n)
{   
    if (n <= 1) return false;
    if (n == 2) return true;
    if (n%2 == 0) return false;

    int sqrtn,i;
    sqrtn = asmSqrt(n);

    for (i = 3; i <= sqrtn; i+=2) if (n%i == 0) return false;
    return true;
}

//number generator iterated from 0 to n
int main()
{
    n = 1000000; //maximum number
    int k,j;

    for (j = 0; j<= n; j++)
    {
        if(isPrime(j) == 1) k++;
        if(j == n) printf("Count: %d\n",k);
    }
    return 0;
}

第一次尝试并行化

我让pthread管理for loop

# include <stdio.h>
.
.

int main()
{
    .
    .
    //----->pthread code here<----
    for (j = 0; j<= n; j++)
    {
        if(isPrime(j) == 1) k++;
        if(j == n) printf("Count: %d\n",k);
    }
    return 0;
}

嗯，它的运行速度比串行的慢

第二次尝试

我将它们for loop分成两个线程并使用并行运行它们pthreads

但是，它的运行速度仍然较慢，我打算将它的运行速度提高两倍或更快。但它不是！

编辑：顺便说一下，这是我的并行代码：

# include <stdio.h>
# include <pthread.h>
# include <cmath>

# define NTHREADS 2

pthread_mutex_t mutex1 = PTHREAD_MUTEX_INITIALIZER;
int k = 0;

double asmSqrt(double x) 
{
  __asm__ ("fsqrt" : "+t" (x));
  return x;
}

struct arg_struct
{
    int initialPrime;
    int nextPrime;
};

bool isPrime(int n)
{   
    if (n <= 1) return false;

    if (n == 2) return true;

    if (n%2 == 0) return false;

    int sqrtn,i;
    sqrtn = asmSqrt(n);

    for (i = 3; i <= sqrtn; i+=2) if (n%i == 0) return false;

    return true;
}

void *parallel_launcher(void *arguments)
{
    struct arg_struct *args = (struct arg_struct *)arguments;

    int j = args -> initialPrime;
    int n = args -> nextPrime - 1;

    for (j = 0; j<= n; j++)
    {
        if(isPrime(j) == 1)
        {
            printf("This is prime: %d\n",j);
pthread_mutex_lock( &mutex1 );
            k++;
pthread_mutex_unlock( &mutex1 );
        }

        if(j == n) printf("Count: %d\n",k);
    }
pthread_exit(NULL);
}

int main()
{
    int f = 100000000;
    int m;

    pthread_t thread_id[NTHREADS];
    struct arg_struct args;

    int rem = (f+1)%NTHREADS;
    int n = floor((f+1)/NTHREADS);

    for(int h = 0; h < NTHREADS; h++)
    {
        if(rem > 0)
        {
            m = n + 1;
            rem-= 1;
        }
        else if(rem == 0)
        {
            m = n;
        }

        args.initialPrime = args.nextPrime;
        args.nextPrime = args.initialPrime + m;

        pthread_create(&thread_id[h], NULL, &parallel_launcher, (void *)&args);
        pthread_join(thread_id[h], NULL);
    }
   // printf("Count: %d\n",k);
    return 0;
}

注意： 操作系统：Fedora 21 x86_64，编译器：gcc-4.4，处理器：Intel Core i5（2 个物理核心，4 个逻辑），内存：6 Gb，HDD：340 Gb，