函数逻辑报告

函数名称：The thing about the wake_up_state() return value; I think we can ignore it.* If for some reason it would return 0, that means the previously waiting* task is already running, so it will observe condition true (or has already).

函数原型：void swake_up_locked(struct swait_queue_head *q)

返回类型：void

参数：