我为此工作,但没有找到解决问题的办法。
这是我的代码。此代码在pthread_create和pthread_join行上给出错误。我尽一切努力解决了这个问题,但是我做不到。
#include <stdio.h>
#include <pthread.h>
#define array_size 1000
#define no_threads 10
float a[array_size];
int global_index = 0;
int sum = 0;
pthread_t mutex_t ,mutex1;
void *slave(void *ignored)
{
int local_index, partial_sum = 0;
do {
pthread_t mutex_t ,lock(mutex1);
local_index = global_index;
global_index++;
pthread_t mutex ,unlock(mutex1);
if (local_index < array_size)
partial_sum += *(a + local_index);
}
while
(local_index < array_size);
pthread_t mutex , lock(mutex1);
sum += partial_sum;
pthread_t mutex_t , unlock(mutex1);
return 0;
}
main()
{
int i;
pthread_t thread_x[10];
pthread_mutex_init(&mutex1, NULL);
for (i = 0; i < array_size; i++)
a[i] = i+1;
for (i = 0; i < no_threads ; i++)
pthread_create(&thread_x[i] , NULL, slave,NULL);
for (i = 0; i < no_threads; i++)
pthread_join(&thread_x[i] , NULL);
printf("The sum of 1 to %d is %d\n", array_size, sum);
}
最佳答案
我已经更新了您的代码(pt1.c),我想您正在寻找类似的东西。
#include <stdio.h>
#include <pthread.h>
#define array_size 1000
#define no_threads 10
float a[array_size];
int global_index = 0;
int sum = 0;
pthread_mutex_t mutex1;
void *slave(void *ignored)
{
int local_index, partial_sum = 0;
do {
pthread_mutex_lock(&mutex1);
local_index = global_index;
global_index++;
pthread_mutex_unlock(&mutex1);
if (local_index < array_size)
partial_sum += *(a + local_index);
}
while(local_index < array_size);
pthread_mutex_lock(&mutex1);
sum += partial_sum;
pthread_mutex_unlock(&mutex1);
return 0;
}
main()
{
int i;
pthread_t thread_x[10];
pthread_mutex_init(&mutex1, NULL);
for (i = 0; i < array_size; i++)
a[i] = i+1;
for (i = 0; i < no_threads ; i++)
pthread_create(&thread_x[i] , NULL, slave,NULL);
for (i = 0; i < no_threads; i++)
pthread_join(thread_x[i] , NULL);
printf("The sum of 1 to %d is %d\n", array_size, sum);
}
您应该将此代码编译为
gcc -pthread pt1.c
输出:
1至1000的总和为500500
关于c - 如何在C中将具有线程的数组元素求和?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/33949085/