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 10:34:52
Last Modify:2020-03-12 14:18:49 Copyright©Brick
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函数名称：may_create_in_sticky - Check whether an O_CREAT open in a sticky directory* should be allowed, or not, on files that already* exist

函数原型：static int may_create_in_sticky(umode_t dir_mode, kuid_t dir_uid, struct inode *const inode)

返回类型：int

参数：

类型参数名称
umode_tdir_mode
kuid_tdir_uid
struct inode *constinode
1024  如果非sysctl_protected_fifos且S_ISFIFO(i_mode)或非sysctl_protected_regular且S_ISREG(i_mode)或此条件成立可能性大(为编译器优化)(!(dir_mode & S_ISVTX))或uid_eq(i_uid, dir_uid)或uid_eq(current_fsuid(), i_uid)则返回:0
1031  如果此条件成立可能性大(为编译器优化)(dir_mode & 0002)或dir_mode按位与0020且sysctl_protected_fifos大于等于2且S_ISFIFO(i_mode)或sysctl_protected_regular大于等于2且S_ISREG(i_mode)的值则
1035  operation等于如果S_ISFIFO(i_mode)则"sticky_create_fifo"否则"sticky_create_regular"
1038  audit_log_path_denied - report a path restriction denial*@type: audit message type (AUDIT_ANOM_LINK, AUDIT_ANOM_CREAT, etc)*@operation: specific operation name
1039  返回:负EACCES
1041  返回:0
调用者
名称描述
do_lastHandle the last step of open()