no_subtree_check是NFS(Network File System)配置中的一个选项,用于控制NFS服务器对客户端文件句柄(file handle)的挂载点变化及其路径上目录访问权限的检查行为。当启用此选项时,NFS服务器将不会检查客户端持有的文件句柄所指向的目录是否在原始的挂载点内,或者从文件句柄指向的目录到挂载点之间的目录是否有访问权限。
"no_subtree_check"这个词通常出现在网络路由配置中,特别是在边界网关协议(BGP)的上下文中。BGP是一种标准化的外部网关协议,用于在互联网上的自治系统(AS)之间交换路由可达信息。 在网络配置中使用"no_subtree_check"时,它改变了路由聚合的行为。路由聚合,也称为路由汇总,是一种技术,将多个具体的路由合并成一个较...
no_subtree_check: 这个选项禁用子树检查。默认情况下,NFS 会进行子树检查以确保文件系统的稳定性和一致性。禁用子树检查可以提高性能,但在某些情况下可能会导致问题,因此需要谨慎使用。 示例解释 假设您希望将本地目录/mydata共享给所有客户端,并且希望客户端具有读写权限、数据同步写入磁盘、以及禁用子树检查,您可以在...
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What is the default /etc/exports value for subtree_check on RHEL5? Solution Verified- UpdatedAugust 7 2024 at 6:50 AM- English Issue It seems there has been a change in the exports manpage between RHEL 5.3 and RHEL 5.4 about the default of the subtree check option for a NFS export. ...
is it possible to allow subtree checking in omnios and if so how do i accomplish this?G gea Well-Known Member Dec 31, 2010 3,414 1,332 113 DE Dec 28, 2016 #2 What do you mean with subtree checking? NFS (like SMB) is a pure ZFS filesystem property on SolarishM...
It returns true if subRoot is a subtree of root and false otherwise. public bool IsSubtree(TreeNode root, TreeNode subRoot) { if (root == null) { return false; } if (CheckSubtree(root, subRoot)) return true; return IsSubtree(root.left, subRoot) || IsSubtree(root.right, subRoot)...
Check subtree is active before adding talent points. Browse files thewarwithin v11.0.5-1.1.0-alpha1 … v11.0.2-1.0.4 Hekili committed Aug 25, 2024 1 parent c75fe36 commit d46b4d9 Showing 1 changed file with 1 addition and 1 deletion. Whitespace Ignore whitespace Split Un...
nfs服务状态显示Assuming default behaviour ('no_subtree_check'). # /etc/init.d/nfs-kernel-server status ● nfs-server.service - NFS server and services Loaded: loaded (/lib/systemd/system/nfs-server.service; enabled; vendor preset: enabled) Active: active (exited) since 三 2019-11-06 11:...
There is a subtree if the given sum is 6 and the subtree is, Of course, the brute force method is to find subtree sum rooted by each node and to check if the subtree sum matches with the input sum. But the complexity would beO(n^2). ...