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 |
|---|---|
| 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 |