函数逻辑报告 |
Source Code:include\linux\oom.h |
Create Date:2022-07-27 09:55:55 |
| 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：tsk_is_oom_victim
函数原型：static inline bool tsk_is_oom_victim(struct task_struct *tsk)
返回类型：bool
参数：
| 类型 | 参数 | 名称 |
|---|---|---|
| struct task_struct * | tsk |
| 76 | 返回:oom_mm |
| 名称 | 描述 |
|---|---|
| __cpuset_node_allowed | puset_node_allowed - Can we allocate on a memory node?*@node: is this an allowed node?*@gfp_mask: memory allocation flags* If we're in interrupt, yes, we can always allocate. If @node is set in* current's mems_allowed, yes |
| oom_evaluate_task | |
| warn_alloc_show_mem | |
| oom_reserves_allowed | |
| __alloc_pages_slowpath | |
| should_force_charge |
| 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 |