pthread_join 死锁的简单示例

Question

我正在寻找一个非常简单的示例来演示使用 pthread_join 的死锁；然而，这并非微不足道。

我是从这个开始的：

void* joinit(void* tid)
{
  pthread_t* tid_c = (pthread_t*)tid;
  int retval = pthread_join(*tid_c, NULL);
  printf("In joinit: tid = %d, retval = %d \n", *tid_c, retval);
  return NULL;
}

int main()
{
  pthread_t thread1;
  pthread_t thread2;

  pthread_create(&thread1, NULL, joinit, &thread2);
  pthread_create(&thread2, NULL, joinit, &thread1);

  pthread_join(thread2, NULL);  

  return 0;
}

但是，它说 'EINVAL'（无效参数），因为在调用 thread1 的 pthread_create 时尚未指定 thread2。

有什么想法吗？

Answer 1

如果您只是想证明 pthread_join 会导致死锁，您可以执行类似于以下代码的操作：

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

void* joinit(void* tid)
{
    printf("In %#x, waiting on %#x\n", pthread_self(), (*((pthread_t*)tid)));
    pthread_join((*((pthread_t*)tid)), NULL);
    printf("Leaving %#x\n", pthread_self());
    return NULL;
}

int main(void)
{
    pthread_t thread1 = pthread_self();
    pthread_t thread2;
    pthread_create(&thread2, NULL, joinit, &thread1);
    joinit(&thread2);
    return 0;
}

这将导致主线程在派生线程上等待，而派生线程在主线程上等待（导致死锁），而不需要额外的锁定原语来扰乱您要演示的内容。

并更直接地回答您的一些问题：

it says 'EINVAL' (invalid argument) because thread2 is not yet specified when pthread_create for thread1 is called.

...以及您的评论之一...

I tried this and it worked, but the problem is, it only works SOMETIMES because sometimes I get EINVAL again.

在您的代码中，您连续调用 pthread_create 以生成 2 个线程：

pthread_create(&thread1, NULL, joinit, &thread2);
pthread_create(&thread2, NULL, joinit, &thread1);

在您的 joinit 代码中，您获取传入的线程句柄以加入：

pthread_t* tid_c = (pthread_t*)tid;
int retval = pthread_join(*tid_c, NULL);

这个有时起作用的原因和你会得到 EINVAL 的其他原因与 time slices allocated to each thread's context and sequencing 有关。当第一个 pthread_create 被调用时，您将在 returns 之后有一个 thread1 的有效句柄，但 thread2 的句柄还无效，至少在第二个 pthread_create 被调用。

为此，当一个线程被创建时，线程的行为"alive"（即线程函数实际上运行）可能需要一些额外的时间，即使返回的线程句柄是有效的.在这些情况下，一个线程执行的代码可能比 "expected" 多。在你的代码中，两个 pthread_create 函数可能恰好在分配给主线程的时间片中被调用，可以给每个在命中 pthread_join 允许 tid_c 指向有效句柄的语句之前生成足够多的线程 "time"；在 EINVAL 的情况下，调用 pthread_create(&thread1, NULL, joinit, &thread2) 并且生成的线程在 pthread_create(&thread2, NULL, joinit, &thread1) 之前命中 pthread_join(*tid_c, NULL) pthread_create(&thread2, NULL, joinit, &thread1) 可以给出 thread2一个有效的句柄（导致错误）。

如果您想让您的代码与现在类似，您需要添加某种锁以确保线程不会退出或过早地调用任何东西：

#include <stdio.h> #include <pthread.h> static pthread_mutex_t lock = PTHREAD_MUTEX_INITIALIZER; void* joinit(void* tid) { /* this can be above the lock because it will be valid after lock is acquired */ pthread_t* tid_c = (pthread_t*)tid; int retval = -1; pthread_mutex_lock(&lock); pthread_mutex_unlock(&lock); printf("%#x waiting on %#x\n", pthread_self(), *tid_c); retval = pthread_join(*tid_c, NULL); printf("In joinit: tid = %d, retval = %d \n", *tid_c, retval); return NULL; } int main() { pthread_t thread1; pthread_t thread2; /* get the lock in the main thread FIRST */ pthread_mutex_lock(&lock); pthread_create(&thread1, NULL, joinit, &thread2); pthread_create(&thread2, NULL, joinit, &thread1); /* by this point, both handles are "joinable", so unlock */ pthread_mutex_unlock(&lock); /* can wait on either thread, but must wait on one so main thread doesn't exit */ pthread_join(thread2, NULL); return 0; }

希望对您有所帮助。

Answer 2

错误的主要原因是您有两个线程，每个线程都在等待同一个线程终止，因为在 main 中调用了 pthread_join。另一个问题是您不能确保每个线程都能正确看到另一个线程的 ID。

像这样修复它：

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

pthread_t thread1;
pthread_t thread2;
pthread_mutex_t mutex;
pthread_cond_t cond;
int go = 0;


void* joinit(void* ptr)
{
  // wait until both thread IDs are known
  pthread_mutex_lock(&mutex);
  while (go == 0)
      pthread_cond_wait(&cond, &mutex);
  pthread_mutex_unlock(&mutex);

  pthread_t* tid_c = *((pthread_t**) ptr);
  printf("About to wait\n");
  int retval = pthread_join(*tid_c, NULL);
  printf("In joinit: tid = %d, retval = %d \n", *tid_c, retval);

  // tell the other threads we're done
  pthread_mutex_lock(&mutex);
  go++;
  pthread_cond_broadcast(&cond);
  pthread_mutex_unlock(&mutex);

  return NULL;
}

int main()
{
  // setup synchronization
  pthread_mutex_init(&mutex, NULL);
  pthread_cond_init(&cond, NULL);

  pthread_create(&thread1, NULL, joinit, &thread2);
  pthread_create(&thread2, NULL, joinit, &thread1);

  // tell the threads to go
  pthread_mutex_lock(&mutex);
  go = 1;
  pthread_cond_broadcast(&cond);
  pthread_mutex_unlock(&mutex);

  // wait for both threads to finish
  pthread_mutex_lock(&mutex);
  while (go != 3)
      pthread_cond_wait(&cond, &mutex);
  pthread_mutex_unlock(&mutex);

  return 0;
}

pthread_join 死锁的简单示例

Simple example for pthread_join deadlock

c

multithreading

deadlock

pthreads

pthread-join