c - Linux 中的 C 线程同步

Question

我在同步线程时遇到问题，因此每个线程可以先运行一个作业，然后另一个线程启动同一个作业，依此类推。下面是我的代码:

#include <unistd.h>
#include <sys/types.h> 
#include <errno.h>
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <string.h>
#include <semaphore.h>

void handler ( void *ptr );
sem_t mutex;

int worker = 2; // number of workers
int job = 4; // number of jobs for each worker

int main()
{

    int i = 0;
    pthread_t thread_a;

    sem_init(&mutex, 0, 1);

    for (i; i < worker; i++)
    {  
       int *n_workers = malloc(sizeof(*n_workers));  
       *n_workers = i;                   
       pthread_create (&thread_a, NULL, (void *) &handler, n_workers);
    }

    pthread_join(thread_a, NULL);

    sem_destroy(&mutex);

    pthread_exit(0);
}

void handler ( void *ptr )
{
    int x = *((int *) ptr);
    int i = 0;

    for (i; i < job; i++)
    {
       sem_wait(&mutex);
       printf("Worker %d: Doing Job %d\n", x, i);  
       sem_post(&mutex);
    }
}

输出是:

Worker 1: Doing Job 0
Worker 1: Doing Job 1
Worker 1: Doing Job 2
Worker 1: Doing Job 3
Worker 0: Doing Job 0
Worker 0: Doing Job 1
Worker 0: Doing Job 2
Worker 0: Doing Job 3

程序中，每个 worker 有4个工作岗位，共有2个 worker 。问题是 worker 1 一次完成所有 4 个工作，而 worker 0 在它之后完成所有工作。理想的输出是这样的:

Worker 0: Doing Job 0
Worker 1: Doing Job 0
Worker 0: Doing Job 1
Worker 1: Doing Job 1
Worker 0: Doing Job 2
Worker 1: Doing Job 2
Worker 0: Doing Job 3
Worker 1: Doing Job 3

我不确定问题出在哪里，非常感谢任何帮助。谢谢

Answer 1

如果你想确保每个线程在作业 [n+1] 之前执行作业 [n]，你需要使用 barriers ，通过使用信号量(为此您可能需要咨询 The Little Book of Semaphores )或使用 pthread_barrier_t 实现它们。

如果您选择后者，您应该能够通过对代码进行最少的修改来实现效果:

#include <unistd.h>
#include <sys/types.h> 
#include <errno.h>
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <string.h>
#include <semaphore.h>

void handler ( void *ptr );
pthread_barrier_t barrier;

int worker = 2; // number of workers
int job = 4; // number of jobs for each worker

int main()
{

    int i = 0;
    pthread_t thread_a;

    pthread_barrier_init(&barrier, NULL, worker);

    for (i; i < worker; i++)
    {  
       int *n_workers = malloc(sizeof(*n_workers));  
       *n_workers = i;                   
       pthread_create (&thread_a, NULL, (void *) &handler, n_workers);
    }

    pthread_join(thread_a, NULL);

    pthread_barrier_destroy(&barrier);

    pthread_exit(0);
}

void handler ( void *ptr )
{
    int x = *((int *) ptr);
    int i = 0;

    for (i; i < job; i++)
    {
       printf("Worker %d: Doing Job %d\n", x, i);  
       pthread_barrier_wait(&barrier);
    }
}

这样，每个作业 [n] 都会在作业 [n+1] 开始之前执行 int worker = 2 次。工作人员在 job[n] 上工作的顺序无关紧要，因此您可能会得到不同的输出，例如:

Worker 0: Doing Job 0
Worker 1: Doing Job 0
Worker 1: Doing Job 1
Worker 0: Doing Job 1
Worker 0: Doing Job 2
Worker 1: Doing Job 2
Worker 0: Doing Job 3
Worker 1: Doing Job 3

重要的是最右边的列。