Question

#include<pthread.h>
#include<stdio.h>
#include<stdlib.h>
#define NUM_THREADS 4

int arraySum = 0;
int array[] = {20, 18, 16, 14, 12, 10, 8, 6, 4, 2, -10, -20, -30, -40, 15, 23, 25, 75, 45, 33};
int low = 0;
int high = 4;

// Function that calculates sum of 5 elements of an array
void *ArraySum(void *threadid)
{
    // Iterating from index low to high
    for(int i = low; i < high; i++)
    {
        // Accumulating array sum
        arraySum = arraySum + array[i];
    }
    // Updating lowest and highest index
    low = low + 5;
    high = high + 5;
    // Exiting current thread
    pthread_exit(NULL);
}
// Main function
int main(int argc, char *argv[])
{
    int rc, t;
    // Creating thread array
    pthread_t threads[NUM_THREADS];
    // Iterating over each thread
    for(t = 0; t < NUM_THREADS; t++)
    {
        // Creating a thread
        rc = pthread_create(&threads[t], NULL, ArraySum, &t);
        // Checking for status
        if(rc)
        {
        printf(" Error; return code from pthread_create() is %d \n", rc);
        exit(-1);
        }
    }
     /* Waiting till all threads finish their execution */
     for (t = 0; t < NUM_THREADS; t++)
     {
         pthread_join(threads[t], NULL);
     }
     // Printing Array sum
     printf("\nArray Sum: %d \n", arraySum);
     pthread_exit(NULL);
 }

目标是创建4个不同的线程并将数组中的所有数字加在一起。例如，第一个线程将添加前5个数字（20 + 18 + 16 + 14 + 12）。第二个线程将添加下一个5，依此类推。

当我运行它时，我总共获得164但我期望211.我相信我创建了4个不同的线程并正确加入它们但输出错误。

Answer 1

在这里，我编写了将不同线程相加的解决方案。我基本上使用线程id来确定每个线程的开始/结束索引。因此，每个线程计算一个部分和，因此不需要同步。当一个线程退出时，他的部分金额将被返回并累积在最终金额中。

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

#define NUM_THREADS 4

int array[] = {20, 18, 16, 14, 12, 10, 8, 6, 4, 2, -10, -20, -30, -40, 15, 23, 25, 75, 45, 33};

// Function that calculates sum of 5 elements of an array
void *ArraySum(void *arg)
{
    int id = (int)arg;
    int partial_sum = 0;
    int begin = id * (NUM_THREADS + 1);
    int end = id * (NUM_THREADS + 1) + NUM_THREADS;

    // Iterating from index low to high
    for(int i = begin; i <= end; i++)
    {
        // Accumulating array partial sum
        partial_sum += array[i];
    }
    // Exiting current thread
    pthread_exit((void *)partial_sum);
}
// Main function
int main(int argc, char *argv[])
{
    int rc, t;
    int arraySum = 0;
    void *retvalue;

    // Creating thread array
    pthread_t threads[NUM_THREADS];

    // Iterating over each thread
    for(t = 0; t < NUM_THREADS; t++)
    {
        // Creating a thread
        rc = pthread_create(&threads[t], NULL, ArraySum, (void *)t);
        // Checking for status
        if(rc)
        {
            printf(" Error; return code from pthread_create() is %d \n", rc);
            exit(-1);
        }
    }
    /* Waiting till all threads finish their execution */
    for (t = 0; t < NUM_THREADS; t++)
    {
         pthread_join(threads[t], &retvalue);
         arraySum += (int)retvalue;
    }
    // Printing Array sum
    printf("\nArray Sum: %d \n", arraySum);
    pthread_exit(NULL);
}

Answer 2

如果您在共享变量（在您的情况下为 arraySum ）上添加互斥锁，您的代码将会很好地工作这是我尝试使用互斥锁的示例代码：

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

#define NUM_THREADS 5
pthread_t threads[NUM_THREADS];
pthread_mutex_t mutexsum;
int  a[2500];
int sum = 0;
void *do_work(void* parms) {

    long tid = (long)parms;
    printf("I am thread # %ld\n ", tid);

    int start, end, mysum;

    start = (int)tid * 500;
    end = start + 500;
    int i = 0;
    printf("Thread # %ld with start = %d and end = %d \n",tid,start,end);
    for (int i = start; i < end; i++) 
    {
        mysum += a[i];
    }
    pthread_mutex_lock(&mutexsum);
    printf("Thread # %ld lock and sum = %d\n",tid,sum);
    sum += mysum;
    pthread_mutex_unlock(&mutexsum);

    pthread_exit(NULL);


    }




void main(int argv, char* argc) {
    int i = 0; int rc;
    pthread_attr_t attr;
    pthread_mutex_init(&mutexsum, NULL);
    pthread_attr_init(&attr);
    pthread_attr_setdetachstate(&attr, PTHREAD_CREATE_JOINABLE);
    pthread_mutex_init(&mutexsum, NULL);
    printf("Initializing array : \n");
    for(i=0;i<2500;i++){
        a[i]=1;
      }
    for (i = 0; i < NUM_THREADS; i++) {
        printf("Creating thread # %d.\n", i);

        rc = pthread_create(&threads[i], &attr, &do_work, (void *)i);
        if (rc) {
            printf("Error in thread %d with rc  = %d. \n", i, rc);
            exit(-1);
        }

    }
    pthread_attr_destroy(&attr);
    printf("Creating threads complete. start ruun " );
    for (i = 0; i < NUM_THREADS; i++) {
        pthread_join(threads[i], NULL);
    }
    printf("\n\tSum : %d", sum);
    pthread_mutex_destroy(&mutexsum);
    pthread_exit(NULL);
}

在C中汇总不同的线程？

2 个答案: