Quoting Dan Smith (danms@xxxxxxxxxx):
> Make this helper available to others.
No objection to exporting may_setuid, nor to creating may_setgid(),
but I don't think may_setgid() is right. See below
> Signed-off-by: Dan Smith <danms@xxxxxxxxxx>
> ---
> include/linux/user.h | 9 +++++++++
> kernel/user.c | 16 +++++++++++++++-
> 2 files changed, 24 insertions(+), 1 deletions(-)
>
> diff --git a/include/linux/user.h b/include/linux/user.h
> index 68daf84..713bae7 100644
> --- a/include/linux/user.h
> +++ b/include/linux/user.h
> @@ -1 +1,10 @@
> +#ifndef _LINUX_USER_H
> +#define _LINUX_USER_H
> +
> #include <asm/user.h>
> +#include <linux/sched.h>
> +
> +extern int may_setuid(struct user_namespace *ns, uid_t uid);
> +extern int may_setgid(struct group_info *groupinfo, gid_t gid);
> +
> +#endif
> diff --git a/kernel/user.c b/kernel/user.c
> index a535ed6..38b8b50 100644
> --- a/kernel/user.c
> +++ b/kernel/user.c
> @@ -604,7 +604,7 @@ int checkpoint_user(struct ckpt_ctx *ctx, void *ptr)
> return do_checkpoint_user(ctx, (struct user_struct *) ptr);
> }
>
> -static int may_setuid(struct user_namespace *ns, uid_t uid)
> +int may_setuid(struct user_namespace *ns, uid_t uid)
> {
> /*
> * this next check will one day become
> @@ -631,6 +631,20 @@ static int may_setuid(struct user_namespace *ns, uid_t uid)
> return 0;
> }
>
> +int may_setgid(struct group_info *groupinfo, gid_t gid)
> +{
> + if (capable(CAP_SETGID))
> + return 1;
We should pass in a user_ns so we can eventually check for
capable_to(ns, CAP_SETGID). So the caller may not be
CAP_SETGID in the root user namespace, but may have created
the child user namespace and be privileged there.
> + if (current_cred_xxx(group_info) != groupinfo)
> + return 0;
I think it's possible for two different group_info's to have the same
member groups.
I think the thing to do is walk over all groups in group_info,
and do in_egroup_p(g) for each.
It's also possible that groups 1, 4, and 5 are in current_group_info and 6 is
current_egroup, while we're asking for a group_info with groups 1, 5 and 6, and
egid of 4. That would be legal, right? Walking over the groups and doing
in_egroup_p(g) should do that check. Sure, it's n^2 on the # groups... So
we could eventually optimize it to exploit the fact that both groupinfos
are sorted and keep last_used_g in both groupinfos...
> + if (in_egroup_p(gid))
> + return 1;
> +
> + return 0;
> +}
> +
> static struct user_struct *do_restore_user(struct ckpt_ctx *ctx)
> {
> struct user_struct *u;
> --
> 1.6.0.4
>
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