函数逻辑报告 |
Source Code:kernel\sched\rt.c |
Create Date:2022-07-27 10:39:42 |
Last Modify:2020-03-12 14:18:49 | Copyright©Brick |
首页 | 函数Tree |
注解内核,赢得工具 | 下载SCCT | English |
函数名称:rt_rq_of_se
函数原型:static inline struct rt_rq *rt_rq_of_se(struct sched_rt_entity *rt_se)
返回类型:struct rt_rq
参数:
类型 | 参数 | 名称 |
---|---|---|
struct sched_rt_entity * | rt_se |
246 | rq等于rq_of_rt_se(rt_se) |
248 | 返回:rt |
名称 | 描述 |
---|---|
update_curr_rt | Update the current task's runtime statistics. Skip current tasks that* are not in our scheduling class. |
__enqueue_rt_entity | |
__dequeue_rt_entity | |
dequeue_rt_stack | Because the prio of an upper entry depends on the lower* entries, we must remove entries top - down. |
requeue_task_rt |
源代码转换工具 开放的插件接口 | X |
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支持:c/c++/esqlc/java Oracle/Informix/Mysql 插件可实现:逻辑报告 代码生成和批量转换代码 |