Question

我有2个线程来创建thread1和thread2。使用以下命令创建线程时

pthread_create(&thread1, NULL, &function_th1, NULL);
pthread_create(&thread2, NULL, &function_th2, NULL);

thread2在thread1之前启动，我正在寻找在thread2之前启动thread1的解决方案。

不寻找这个解决方案：

pthread_create(&thread2, NULL, &function_th2, NULL);
pthread_create(&thread1, NULL, &function_th1, NULL);

Answer 1

发出线程创建调用和线程实际开始执行之间没有关系。这一切都取决于实现，操作系统等，这就是为什么你看到两个线程实际上以看似随机的顺序开始。

如果需要线程1在线程2之前启动，则可能对某些事件有一些数据/逻辑依赖性。在这种情况下，您应该使用条件变量来告诉线程2何时实际开始执行。

// condition variable
pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
// associated mutex
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
// state for condition variable
int shouldWait = 1;

...

void* function_th1(void* p) {
     // execute something

     // signal thread 2 to proceed
     pthread_mutex_lock(&mutex);
     shouldWait = 0;
     pthread_cond_signal(&cond);
     pthread_mutex_unlock(&mutex);

     // other stuff
}

void* function_th2(void* p) {
     // wait for signal from thread 1
     pthread_mutex_lock(&mutex);
     while(shouldWait) {
         pthread_cond_wait(&cond, &mutex);
     }
     pthread_mutex_unlock(&mutex);

     // other stuff
}

Answer 2

移动'pthread_create（＆amp; thread2，NULL，＆amp; function_th2，NULL）;'到'function_th1'函数的顶部。

这肯定会实现你的要求，不需要复杂的信号。

你要求的是你真正想要或需要的是另一回事。

Answer 3

我建议的解决方案之后

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

static pthread_mutex_t mut = PTHREAD_MUTEX_INITIALIZER;
static pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
static bool wait = TRUE;

void thread_sync() {
  pthread_mutex_lock(&mut);
  wait = FALSE;
  pthread_cond_signal(&cond);
  pthread_mutex_unlock(&mut);
}
void thread_wait_sync() {
  pthread_mutex_lock(&mut);
  if (wait==TRUE)
  {
      pthread_cond_wait(&cond,&mut);
  }
  wait = TRUE;
  pthread_mutex_unlock(&mut);
}

void *thread1(void *d) {
  thread_sync();
  while (1); // Rest of work happens whenever
  return NULL;
}

void *thread2(void *d) {
  thread_sync();
  while (1);
  return NULL;
}

void *thread3(void *d) {
  thread_sync();
  while (1);
  return NULL;
}

void *thread4(void *d) {
  while (1);
  return NULL;
}

int main() {
  pthread_t t1,t2,t3,t4;
  pthread_create(&t1, NULL, thread1, NULL);
  thread_wait_sync();
  pthread_create(&t2, NULL, thread2, NULL);
  thread_wait_sync();
  pthread_create(&t3, NULL, thread3, NULL);
  thread_wait_sync();
  pthread_create(&t4, NULL, thread4, NULL);
  while(1) {
    // some work
  }
}

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3 个答案: