函数逻辑报告 |
Source Code:mm\zbud.c |
Create Date:2022-07-27 18:00:32 |
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:Converts an allocation size in bytes to size in zbud chunks
函数原型:static int size_to_chunks(size_t size)
返回类型:int
参数:
类型 | 参数 | 名称 |
---|---|---|
size_t | size |
232 | 返回:size加CHUNK_SIZE减1右移CHUNK_SHIFT位 |
名称 | 描述 |
---|---|
zbud_alloc | zbud_alloc() - allocates a region of a given size*@pool: zbud pool from which to allocate*@size: size in bytes of the desired allocation*@gfp: gfp flags used if the pool needs to grow*@handle: handle of the new allocation* This function will attempt to |
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 |