函数逻辑报告 |
Source Code:mm\zbud.c |
Create Date:2022-07-27 18:00:35 |
Last Modify:2020-03-12 14:18:49 | Copyright©Brick |
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注解内核,赢得工具 | 下载SCCT | English |
函数名称:Returns the number of free chunks in a zbud page
函数原型:static int num_free_chunks(struct zbud_header *zhdr)
返回类型:int
参数:
类型 | 参数 | 名称 |
---|---|---|
struct zbud_header * | zhdr |
292 | 返回:NCHUNKS减first_chunks减last_chunks |
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 |
zbud_free | zbud_free() - frees the allocation associated with the given handle*@pool: pool in which the allocation resided*@handle: handle associated with the allocation returned by zbud_alloc()* In the case that the zbud page in which the allocation resides is |
zbud_reclaim_page | zbud_reclaim_page() - evicts allocations from a pool page and frees it*@pool: pool from which a page will attempt to be evicted*@retries: number of pages on the LRU list for which eviction will* be attempted before failing* zbud reclaim is different from |
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 |