mnt: expose pointer to init_mnt_ns

There's various scenarios where we need to know whether we are in the
initial set of namespaces or not to e.g., shortcut permission checking.
All namespaces expose that information. Let's do that too.

Reviewed-by: Jan Kara <jack@suse.cz>
Signed-off-by: Christian Brauner <brauner@kernel.org>

Author: brauner

fs/namespace.c

@@ -6008,35 +6008,40 @@ SYSCALL_DEFINE4(listmount, const struct mnt_id_req __user *, req,
 	return ret;
 }
 
+struct mnt_namespace init_mnt_ns = {
+	.ns.inum	= PROC_MNT_INIT_INO,
+	.ns.ops		= &mntns_operations,
+	.user_ns	= &init_user_ns,
+	.ns.count	= REFCOUNT_INIT(1),
+	.passive	= REFCOUNT_INIT(1),
+	.mounts		= RB_ROOT,
+	.poll		= __WAIT_QUEUE_HEAD_INITIALIZER(init_mnt_ns.poll),
+};
+
 static void __init init_mount_tree(void)
 {
 	struct vfsmount *mnt;
 	struct mount *m;
-	struct mnt_namespace *ns;
 	struct path root;
 
 	mnt = vfs_kern_mount(&rootfs_fs_type, 0, "rootfs", NULL);
 	if (IS_ERR(mnt))
 		panic("Can't create rootfs");
 
-	ns = alloc_mnt_ns(&init_user_ns, true);
-	if (IS_ERR(ns))
-		panic("Can't allocate initial namespace");
-	ns->ns.inum = PROC_MNT_INIT_INO;
 	m = real_mount(mnt);
-	ns->root = m;
-	ns->nr_mounts = 1;
-	mnt_add_to_ns(ns, m);
-	init_task.nsproxy->mnt_ns = ns;
-	get_mnt_ns(ns);
+	init_mnt_ns.root = m;
+	init_mnt_ns.nr_mounts = 1;
+	mnt_add_to_ns(&init_mnt_ns, m);
+	init_task.nsproxy->mnt_ns = &init_mnt_ns;
+	get_mnt_ns(&init_mnt_ns);
 
 	root.mnt = mnt;
 	root.dentry = mnt->mnt_root;
 
 	set_fs_pwd(current->fs, &root);
 	set_fs_root(current->fs, &root);
 
-	ns_tree_add(ns);
+	ns_tree_add(&init_mnt_ns);
 }
 
 void __init mnt_init(void)

include/linux/mnt_namespace.h

@@ -11,6 +11,8 @@ struct fs_struct;
 struct user_namespace;
 struct ns_common;
 
+extern struct mnt_namespace init_mnt_ns;
+
 extern struct mnt_namespace *copy_mnt_ns(unsigned long, struct mnt_namespace *,
 		struct user_namespace *, struct fs_struct *);
 extern void put_mnt_ns(struct mnt_namespace *ns);