检查 Linux 中 struct mutex 的所有者字段

Question

我想检查一个锁定在内核中的 struct mutex，谁是那个互斥量的所有者。

struct mutex {
    atomic_long_t       owner;
    spinlock_t      wait_lock;
#ifdef CONFIG_MUTEX_SPIN_ON_OWNER
    struct optimistic_spin_queue osq; /* Spinner MCS lock */
#endif
    struct list_head    wait_list;
#ifdef CONFIG_DEBUG_MUTEXES
    void            *magic;
#endif
#ifdef CONFIG_DEBUG_LOCK_ALLOC
    struct lockdep_map  dep_map;
#endif
};

我从 kernel documentation 了解到：

Field owner actually contains struct task_struct * to the current lock owner and it is therefore NULL if not currently owned.

是否有任何安全的方法将该字段与 current 进行比较？

Answer 1

Is there any way to compare that field to current?

从kernel/locking/mutex.c开始你可以使用这个功能：

/*
 * Internal helper function; C doesn't allow us to hide it :/
 *
 * DO NOT USE (outside of mutex code).
 */
static inline struct task_struct *__mutex_owner(struct mutex *lock)
{
    return (struct task_struct *)(atomic_long_read(&lock->owner) & ~MUTEX_FLAGS);
}

那我猜：

extern struct mutex *some_mutex;
extern struct task_struct *current;
__mutex_owner(some_mutex) == current;

Answer 2

如果你想实现递归互斥锁，那么你可以使用mutex_trylock_recursive函数：它的行为类似于mutex_trylock，但是互斥锁的当前所有者允许调用这个函数。在那种情况下，它 returns 一个特殊值。

与__mutex_owner一样，函数mutex_trylock_recursive被标记为已弃用/"do not use"。可能有人不喜欢内核中递归互斥锁的想法。

检查 Linux 中 struct mutex 的所有者字段

Checking the owner field of struct mutex in Linux

c

mutex

linux-kernel