C中读写锁的线程安全销毁
Thread safe destruction of Read-Write Lock in C
我正在尝试使用 POSIX 信号量在 C 语言中编写一个线程安全的读写锁。您可以看到源代码的当前状态here。
我按照 this 创建了读者首选锁。
问题是我想处理调用 rwl_destroy() 时可能处于的任何可能状态的锁销毁。
如果调用 destroy 并且没有其他线程在锁上,那么它将锁定 wrt(由编写器使用)以防止任何其他线程访问锁保护的数据。接下来 destroy 函数应该销毁信号量并释放为 ReadWriteLock 结构分配的内存。但是如果另一个线程正在等待锁呢?根据文档,该线程将处于未定义状态。
这就是我试图避免的,以便使锁更易于使用。
编辑:
当前代码是:
typedef struct ReadWriteLock
{
sem_t wrt;
sem_t mtx;
sem_t delFlag;
int readcount;
int active;
}ReadWriteLock;
//forward declaration
/* This function is used to take the state of the lock.
* Return values:
* [*] 1 is returned when the lock is alive.
* [*] 0 is returned when the lock is marked for delete.
* [*] -1 is returned if an error was encountered.
*/
int isActive(ReadWriteLock*);
int rwl_init(ReadWriteLock* lock)
{
lock = malloc(sizeof(ReadWriteLock));
if (lock == NULL)
{
perror("rwl_init - could not allocate memory for lock\n");
return -1;
}
if (sem_init(&(lock->wrt), 0, 1) == -1)
{
perror("rwl_init - could not allocate wrt semaphore\n");
free(lock);
lock = NULL;
return -1;
}
if (sem_init(&(lock->mtx), 0, 1) == -1)
{
perror("rwl_init - could not allocate mtx semaphore\n");
sem_destroy(&(lock->wrt));
free(lock);
lock = NULL;
return -1;
}
if (sem_init(&(lock->delFlag), 0 , 1) == -1)
{
perror("rwl_init - could not allocate delFlag semaphore\n");
sem_destroy(&(lock->wrt));
sem_destroy(&(lock->mtx));
free(lock);
lock = NULL;
return -1;
}
lock->readcount = 0;
lock->active = 1;
return 0;
}
int rwl_destroy(ReadWriteLock* lock)
{
errno = 0;
if (sem_trywait(&(lock->wrt)) == -1)
perror("rwl_destroy - trywait on wrt failed.");
if ( errno == EAGAIN)
perror("rwl_destroy - wrt is locked, undefined behaviour.");
errno = 0;
if (sem_trywait(&(lock->mtx)) == -1)
perror("rwl_destroy - trywait on mtx failed.");
if ( errno == EAGAIN)
perror("rwl_destroy - mtx is locked, undefined behaviour.");
if (sem_destroy(&(lock->wrt)) == -1)
perror("rwl_destroy - destroy wrt failed");
if (sem_destroy(&(lock->mtx)) == -1)
perror("rwl_destroy - destroy mtx failed");
if (sem_destroy(&(lock->delFlag)) == -1)
perror("rwl_destroy - destroy delFlag failed");
free(lock);
lock = NULL;
return 0;
}
int isActive(ReadWriteLock* lock)
{
errno = 0;
if (sem_trywait(&(lock->delFlag)) == -1)
{
perror("isActive - trywait on delFlag failed.");
return -1;
}
if ( errno == EAGAIN)
{//delFlag is down, lock is marked for delete
perror("isActive - tried to lock but ReadWriteLock was marked for delete");
return 0;
}
return 1;
}
我还有这些功能:
int rwl_writeLock(ReadWriteLock*);
int rwl_writeUnlock(ReadWriteLock*);
int rwl_readLock(ReadWriteLock*);
int rwl_readUnlock(ReadWriteLock*);
所以我的问题是如何更改这些函数以避免出现我上面描述的未定义状态。在尝试销毁 ReadWriteLock 之前,是否有可能或此代码的用户应该负责释放所有锁?
目前没有使用isActive()函数和delFlag信号量,它们是我在尝试解决问题时创建的。
您应该实现 ReadWriteLock 实例的 "disposed" 状态("active" 字段看起来合适,但您不使用它,为什么?)。
在 rwl_writeLock / rwl_readLock 调用 sem_wait() 之前和之后检查两次。这个技巧被称为"double-checking lock pattern"。如果你在进入sem_wait之前发现你的Lock被删除了,就离开这个功能吧。
如果你输入sem_wait后发现你的Lock被删除了,请立即sem_post离开。
在您的 destroy() 例程中,设置 active=0,然后将 sem_post 设置为两个信号量(如果 sem_post 失败,请不要打扰)。如果您之后仍然需要 sem_destroy,请睡一会儿(这样所有的读者和作者都有时间接收信号)并执行 sem_destroy。
P.S。如果您确定不再使用信号量,则实际上无需调用 sem_destroy。
我正在尝试使用 POSIX 信号量在 C 语言中编写一个线程安全的读写锁。您可以看到源代码的当前状态here。 我按照 this 创建了读者首选锁。
问题是我想处理调用 rwl_destroy() 时可能处于的任何可能状态的锁销毁。
如果调用 destroy 并且没有其他线程在锁上,那么它将锁定 wrt(由编写器使用)以防止任何其他线程访问锁保护的数据。接下来 destroy 函数应该销毁信号量并释放为 ReadWriteLock 结构分配的内存。但是如果另一个线程正在等待锁呢?根据文档,该线程将处于未定义状态。
这就是我试图避免的,以便使锁更易于使用。
编辑:
当前代码是:
typedef struct ReadWriteLock
{
sem_t wrt;
sem_t mtx;
sem_t delFlag;
int readcount;
int active;
}ReadWriteLock;
//forward declaration
/* This function is used to take the state of the lock.
* Return values:
* [*] 1 is returned when the lock is alive.
* [*] 0 is returned when the lock is marked for delete.
* [*] -1 is returned if an error was encountered.
*/
int isActive(ReadWriteLock*);
int rwl_init(ReadWriteLock* lock)
{
lock = malloc(sizeof(ReadWriteLock));
if (lock == NULL)
{
perror("rwl_init - could not allocate memory for lock\n");
return -1;
}
if (sem_init(&(lock->wrt), 0, 1) == -1)
{
perror("rwl_init - could not allocate wrt semaphore\n");
free(lock);
lock = NULL;
return -1;
}
if (sem_init(&(lock->mtx), 0, 1) == -1)
{
perror("rwl_init - could not allocate mtx semaphore\n");
sem_destroy(&(lock->wrt));
free(lock);
lock = NULL;
return -1;
}
if (sem_init(&(lock->delFlag), 0 , 1) == -1)
{
perror("rwl_init - could not allocate delFlag semaphore\n");
sem_destroy(&(lock->wrt));
sem_destroy(&(lock->mtx));
free(lock);
lock = NULL;
return -1;
}
lock->readcount = 0;
lock->active = 1;
return 0;
}
int rwl_destroy(ReadWriteLock* lock)
{
errno = 0;
if (sem_trywait(&(lock->wrt)) == -1)
perror("rwl_destroy - trywait on wrt failed.");
if ( errno == EAGAIN)
perror("rwl_destroy - wrt is locked, undefined behaviour.");
errno = 0;
if (sem_trywait(&(lock->mtx)) == -1)
perror("rwl_destroy - trywait on mtx failed.");
if ( errno == EAGAIN)
perror("rwl_destroy - mtx is locked, undefined behaviour.");
if (sem_destroy(&(lock->wrt)) == -1)
perror("rwl_destroy - destroy wrt failed");
if (sem_destroy(&(lock->mtx)) == -1)
perror("rwl_destroy - destroy mtx failed");
if (sem_destroy(&(lock->delFlag)) == -1)
perror("rwl_destroy - destroy delFlag failed");
free(lock);
lock = NULL;
return 0;
}
int isActive(ReadWriteLock* lock)
{
errno = 0;
if (sem_trywait(&(lock->delFlag)) == -1)
{
perror("isActive - trywait on delFlag failed.");
return -1;
}
if ( errno == EAGAIN)
{//delFlag is down, lock is marked for delete
perror("isActive - tried to lock but ReadWriteLock was marked for delete");
return 0;
}
return 1;
}
我还有这些功能:
int rwl_writeLock(ReadWriteLock*);
int rwl_writeUnlock(ReadWriteLock*);
int rwl_readLock(ReadWriteLock*);
int rwl_readUnlock(ReadWriteLock*);
所以我的问题是如何更改这些函数以避免出现我上面描述的未定义状态。在尝试销毁 ReadWriteLock 之前,是否有可能或此代码的用户应该负责释放所有锁?
目前没有使用isActive()函数和delFlag信号量,它们是我在尝试解决问题时创建的。
您应该实现 ReadWriteLock 实例的 "disposed" 状态("active" 字段看起来合适,但您不使用它,为什么?)。
在 rwl_writeLock / rwl_readLock 调用 sem_wait() 之前和之后检查两次。这个技巧被称为"double-checking lock pattern"。如果你在进入sem_wait之前发现你的Lock被删除了,就离开这个功能吧。 如果你输入sem_wait后发现你的Lock被删除了,请立即sem_post离开。
在您的 destroy() 例程中,设置 active=0,然后将 sem_post 设置为两个信号量(如果 sem_post 失败,请不要打扰)。如果您之后仍然需要 sem_destroy,请睡一会儿(这样所有的读者和作者都有时间接收信号)并执行 sem_destroy。
P.S。如果您确定不再使用信号量,则实际上无需调用 sem_destroy。