從 function 指針調用 function 而不分配？

Question

通常我們必須這樣做才能從 function 指針調用 function：

int foo()
{
}

int main()
{
    int (*pFoo)() = foo; // pFoo points to function foo()
    foo();
    return 0;
}

在Linux kernel代碼中， sched_class有很多function個指針：

struct sched_class {
        const struct sched_class *next;

        void (*enqueue_task) (struct rq *rq, struct task_struct *p, int flags);
        void (*dequeue_task) (struct rq *rq, struct task_struct *p, int flags);
        void (*yield_task) (struct rq *rq);
        bool (*yield_to_task) (struct rq *rq, struct task_struct *p, bool preempt);
        .....
}

在pick_next_task function中定義了一個名為class的sched_class的本地實例，直接調用其中的function而不賦值給具有相同簽名的外部函數（從for_each_class開始）：

static inline struct task_struct *
pick_next_task(struct rq *rq)
{
        const struct sched_class *class;
        struct task_struct *p;
    /*
     * Optimization: we know that if all tasks are in
     * the fair class we can call that function directly:
     */
    if (likely(rq->nr_running == rq->cfs.h_nr_running)) {
            p = fair_sched_class.pick_next_task(rq);
            if (likely(p))
                    return p;
    }

    for_each_class(class) {
            p = class->pick_next_task(rq);
            if (p)
                    return p;
    }

    BUG(); /* the idle class will always have a runnable task */
}

是不是因為sched_class中的每一個function指針都與實際實現的sched_class同名，所以每次通過sched_class的sched_class指針調用時，都會自動在kernel地址空間中找到匹配的符號？

Answer 1

for_each_class的定義應該為你清除它

 #define for_each_class(class) \
       for (class = sched_class_highest; class; class = class->next)

如果你 go 跟蹤， sched_class_highest結束這樣的事情

#define sched_class_highest (&stop_sched_class)
extern const struct sched_class stop_sched_class;

/*
* Simple, special scheduling class for the per-CPU stop tasks:
*/
const struct sched_class stop_sched_class = {
      .next                   = &rt_sched_class,

      .enqueue_task           = enqueue_task_stop,
      .dequeue_task           = dequeue_task_stop,
      .yield_task             = yield_task_stop,

      .check_preempt_curr     = check_preempt_curr_stop,

      .pick_next_task         = pick_next_task_stop,
      .put_prev_task          = put_prev_task_stop,

#ifdef CONFIG_SMP
      .select_task_rq         = select_task_rq_stop,
#endif

     .set_curr_task          = set_curr_task_stop,
     .task_tick              = task_tick_stop,

     .get_rr_interval        = get_rr_interval_stop,

     .prio_changed           = prio_changed_stop,
     .switched_to            = switched_to_stop,
};

現在你快樂嗎？ :)

Answer 2

https://github.com/torvalds/linux/blob/v3.3/kernel/sched/sched.h#L850

查看for_each_class宏的擴展。 它在使用之前將值分配給class指針。

Answer 3

每個sched_class結構及其包含的 function 指針都已初始化（否則，它可能是一個錯誤）。 例如，公平調度class在kernel/sched/fair.c中初始化（參見此處）：

const struct sched_class fair_sched_class = {
        .next                   = &idle_sched_class,
        /* lots of assignments */
        .pick_next_task         = pick_next_task_fair,
        /* etc. */
};