Re: [PATCH v2 2/5] seccomp: make underlying bpf ref counted as well

[Daniel Borkmann]

On 09/11/2015 02:20 AM, Tycho Andersen wrote:
> In the next patch, we're going to add a way to access the underlying
> filters via bpf fds. This means that we need to ref-count both the
> struct seccomp_filter objects and the struct bpf_prog objects separately,
> in case a process dies but a filter is still referred to by another
> process.
>
> Additionally, we mark classic converted seccomp filters as seccomp eBPF
> programs, since they are a subset of what is supported in seccomp eBPF.
>
> Signed-off-by: Tycho Andersen <tycho.andersen@xxxxxxxxxxxxx>
> CC: Kees Cook <keescook@xxxxxxxxxxxx>
> CC: Will Drewry <wad@xxxxxxxxxxxx>
> CC: Oleg Nesterov <oleg@xxxxxxxxxx>
> CC: Andy Lutomirski <luto@xxxxxxxxxxxxxx>
> CC: Pavel Emelyanov <xemul@xxxxxxxxxxxxx>
> CC: Serge E. Hallyn <serge.hallyn@xxxxxxxxxx>
> CC: Alexei Starovoitov <ast@xxxxxxxxxx>
> CC: Daniel Borkmann <daniel@xxxxxxxxxxxxx>
> ---
>   kernel/seccomp.c | 4 +++-
>   1 file changed, 3 insertions(+), 1 deletion(-)
>
> diff --git a/kernel/seccomp.c b/kernel/seccomp.c
> index 245df6b..afaeddf 100644
> --- a/kernel/seccomp.c
> +++ b/kernel/seccomp.c
> @@ -378,6 +378,8 @@ static struct seccomp_filter *seccomp_prepare_filter(struct sock_fprog *fprog)
>   	}
>
>   	atomic_set(&sfilter->usage, 1);
> +	atomic_set(&sfilter->prog->aux->refcnt, 1);
> +	sfilter->prog->type = BPF_PROG_TYPE_SECCOMP;

So, if you do this, then this breaks the assumption of eBPF JITs
that, currently, all classic converted BPF programs always have a
prog->type of BPF_PROG_TYPE_UNSPEC (see: bpf_prog_was_classic()).

Currently, JITs make use of this information to determine whether
A and X mappings for such programs should or should not be cleared
in the prologue (s390 currently).

In the seccomp_prepare_filter() stage, we're already past that, so
it will not cause an issue, but we certainly would need to be very
careful in future, if bpf_prog_was_classic() is then used at a later
stage when we already have a generated bpf_prog somewhere, as then
this assumption will break.

>   	return sfilter;
>   }
> @@ -470,7 +472,7 @@ void get_seccomp_filter(struct task_struct *tsk)
>   static inline void seccomp_filter_free(struct seccomp_filter *filter)
>   {
>   	if (filter) {
> -		bpf_prog_free(filter->prog);
> +		bpf_prog_put(filter->prog);
>   		kfree(filter);
>   	}
>   }

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