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mnt: expose pointer to init_mnt_ns

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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>
This commit is contained in:
Christian Brauner
2025-09-17 12:28:01 +02:00
parent 93f67a7dda
commit b2a0b19208
2 changed files with 18 additions and 11 deletions
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27
fs/namespace.c
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@@ -6008,27 +6008,32 @@ 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;
@@ -6036,7 +6041,7 @@ static void __init init_mount_tree(void)
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)