C 如何强制一个线程先于另一个线程启动？

例如，我想创建5个线程并打印它们。如何使第四个在第二个之前执行？我试着用互斥锁锁定它，但我不知道如何只锁定第二个，所以它给了我一个错误

char name1[]="THREAD1";
char name2[]="THREAD2";
char name3[]="THREAD3";
char name4[]="THREAD4";
char name5[]="THREAD5";

pthread_mutex_t mutex;

pthread_t t1, t2,t3,t4,t5;
int* state[5]={0,0,0,0,0};

void* execThread(void* threadName)
{
int i;
int* number = (int*) malloc(sizeof(int));
char* tmp;

if(*(state[4]) == 0 && (pthread_self()==t2))
 pthread_mutex_lock (&mutex);
else if(*(state[4]) == 0 && (pthread_self()!=t2))
{

pthread_yield();
printf("[Thread %s] Executing step %d\n", (char *) threadName, i);

tmp = & ((char*)threadName)[strlen((char*)threadName)-1];
*number = atoi(tmp);

return number;
}
else
{
    pthread_mutex_unlock (&mutex);

    pthread_yield();
    printf("[Thread %s] Executing step %d\n", (char *) threadName, i);


    tmp = & ((char*)threadName)[strlen((char*)threadName)-1];
    *number = atoi(tmp);

    return number;
}
}


void main()
{

pthread_mutex_init(&mutex, NULL);


printf("[Thread MAIN] Starting. tid=%ld, pid=%d\n", (long int) pthread_self(), (int) getpid());

if (pthread_create(&t1, NULL, execThread, name1)!=0) {
perror("Error creating a new thread");
exit(1);
}

if (pthread_create(&t2, NULL, execThread, name2)!=0) {
perror("Error creating a new thread");
exit(1);
}    

if (pthread_create(&t3, NULL, execThread, name3)!=0) {
perror("Error creating a new thread");
exit(1);
} 
if (pthread_create(&t4, NULL, execThread, name4)!=0) {
perror("Error creating a new thread");
exit(1);
} 
if (pthread_create(&t5, NULL, execThread, name5)!=0) {
perror("Error creating a new thread");
exit(1);
} 


printf("[Thread MAIN] Created 5 threads \n" );

pthread_join(t1, (void**) &state[0]);
pthread_join(t2, (void**) &state[1]);
pthread_join(t3, (void**) &state[2]);
pthread_join(t4, (void**) &state[3]);
pthread_join(t5, (void**) &state[4]);

printf("[Thread MAIN] Thread %s[%ld] terminated with state %d\n", name1, (long int) t1, *(state[0]));
printf("[Thread MAIN] Thread %s[%ld] terminated with state %d\n", name2, (long int) t2, *(state[1]));
printf("[Thread MAIN] Thread %s[%ld] terminated with state %d\n", name3, (long int) t3, *(state[2]));
printf("[Thread MAIN] Thread %s[%ld] terminated with state %d\n", name4, (long int) t4, *(state[3]));
printf("[Thread MAIN] Thread %s[%ld] terminated with state %d\n", name5, (long int) t5, *(state[4]));

printf("[Thread MAIN] Terminating. tid=%ld, pid=%d\n", (long int) pthread_self(), (int) getpid());


 pthread_mutex_destroy(&mutex);
//pthread_exit(NULL);
}

---

如果希望线程2仅在线程4打印其消息后打印其消息，则可以使用标志和条件变量。请注意，如果没有进一步的同步，您无法安全地访问线程中的

t1

、

t2

等变量（它们由主线程写入，您试图从其他线程读取）

---

在创建任何线程之前锁定互斥锁。使用更好的机制告诉线程哪个是哪个-将线程号作为（或其中一部分）信息传递给线程函数。第二个线程等待锁定互斥锁；在解锁互斥锁并让第二个线程继续之前，第四个线程执行它需要执行的任何操作。现在还不清楚线程4是第一件事，还是最后一件事，或者介于两者之间。线程1、3、5根本不需要担心互斥锁。在Linux系统上，

void main（）

是无条件错误的。与问题完全相同的问题和代码片段，尽管该代码片段已从另一个问题中删除。这里发生了什么？

int t4_done = 0;
pthread_mutex_t t4_done_mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t t4_done_cond = PTHREAD_COND_INITIALIZER;

void* execThread(void* threadName)
{
    if (threadName == name2)
    {
        /* thread2 waits for thread4 to set the t4_done flag */
        pthread_mutex_lock(&t4_done_mutex);
        while (!t4_done)
            pthread_cond_wait(&t4_done_cond, &t4_done_mutex);
        pthread_mutex_unlock(&t4_done_mutex);
    }

    printf("[Thread %s] Executing step\n", (char *) threadName);

    if (threadName == name4)
    {
        /* thread4 sets the t4_done flag and signals thread2 */
        pthread_mutex_lock(&t4_done_mutex);
        t4_done = 1;
        pthread_cond_signal(&t4_done_cond);
        pthread_mutex_unlock(&t4_done_mutex);
    }

    return NULL;
}