函数逻辑报告

Linux Kernel

v5.5.9

Brick Technologies Co., Ltd

Source Code:fs\namei.c Create Date:2022-07-29 10:34:56
Last Modify:2020-03-12 14:18:49 Copyright©Brick
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函数名称：pick_link

函数原型：static int pick_link(struct nameidata *nd, struct path *link, struct inode *inode, unsigned seq)

返回类型：int

参数：

类型参数名称
struct nameidata *nd
struct path *link
struct inode *inode
unsignedseq
1717  如果此条件成立可能性小(为编译器优化)(total_link_count++ >= MAXSYMLINKS)则
1718  path_to_nameidata(link, nd)
1719  返回:负ELOOP
1721  如果非flags按位与RCU pathwalk mode; semi-internal 的值则
1722  如果mnt恒等于mnt则mntget(mnt)
1725  error等于nd_alloc_stack(nd)
1726  如果此条件成立可能性小(为编译器优化)(error)则
1727  如果error恒等于负ECHILD则
1728  如果此条件成立可能性小(为编译器优化)(!path_put is needed afterwards regardless of success or failure )则
1735  否则如果此条件成立可能性大(为编译器优化)(lazy_walk - try to switch to ref-walk mode)恒等于0则error等于nd_alloc_stack(nd)
1738  如果error则
1739  path_put - put a reference to a path*@path: path to put the reference to* Given a path decrement the reference count to the dentry and the vfsmount.
1740  返回:error
1744  last等于stack加depth自加
1745  link等于link
1746  clear_delayed_call( & done)
1747  link_inode等于inode
1748  seq等于seq
1749  返回:1
调用者
名称描述
step_intoDo we need to follow links? We _really_ want to be able* to do this check without having to look at inode->i_op,* so we keep a cache of "no, this doesn't need follow_link"* for the common case.