On Mon, Oct 16, 2017 at 2:10 PM, Richard Weinberger <rich...@nod.at> wrote: > Am Montag, 16. Oktober 2017, 23:02:06 CEST schrieb Daniel Borkmann: >> On 10/16/2017 10:55 PM, Richard Weinberger wrote: >> > Am Montag, 16. Oktober 2017, 22:50:43 CEST schrieb Daniel Borkmann: >> >>> struct task_struct *task = current; >> >>> >> >>> + task_lock(task); >> >>> >> >>> strncpy(buf, task->comm, size); >> >>> >> >>> + task_unlock(task); >> >> >> >> Wouldn't this potentially lead to a deadlock? E.g. you attach yourself >> >> to task_lock() / spin_lock() / etc, and then the BPF prog triggers the >> >> bpf_get_current_comm() taking the lock again ... >> > >> > Yes, but doesn't the same apply to the use case when I attach to strncpy() >> > and run bpf_get_current_comm()? >> >> You mean due to recursion? In that case trace_call_bpf() would bail out >> due to the bpf_prog_active counter. > > Ah, that's true. > So, when someone wants to use bpf_get_current_comm() while tracing task_lock, > we have a problem. I agree. > On the other hand, without locking the function may return wrong results.
it will surely race with somebody else setting task comm and it's fine. all of bpf tracing is read-only, so locks are only allowed inside bpf core bits like maps. Taking core locks like task_lock() is quite scary. bpf scripts rely on bpf_probe_read() of all sorts of kernel fields so reading comm here w/o lock is fine.
