Re: [PATCH v2 10/11] perf/uprobe: Convert single-step and uretprobe to SRCU

[Peter Zijlstra]

On Thu, Jul 11, 2024 at 06:06:53PM +0200, Oleg Nesterov wrote:
> I'll try to actually apply the whole series and read the code tomorrow.
> Right now I can't understand this change... Just one question for now.
> 
> On 07/11, Peter Zijlstra wrote:
> >
> > @@ -1956,11 +1960,13 @@ static void prepare_uretprobe(struct upr
> >  			 * attack from user-space.
> >  			 */
> >  			uprobe_warn(current, "handle tail call");
> > -			goto err_uprobe;
> > +			goto err_mem;
> >  		}
> >  		orig_ret_vaddr = utask->return_instances->orig_ret_vaddr;
> >  	}
> >
> > +	ri->srcu_idx = __srcu_read_lock(&uretprobes_srcu);
> > +	ri->uprobe = uprobe;
> 
> It seems that, if we race with _unregister, this __srcu_read_lock()
> can happen after call_srcu(uprobes_srcu, uprobe, uprobe_free_stage1)
> was already called...
> 
> In this case read_lock "has no effect" in that uprobe_free_stage1()
> can run before free_ret_instance() does srcu_read_unlock(ri->srcu_idx).
> 
> Perhaps it is fine, uprobe_free_stage1() does another call_srcu(),
> but somehow I got lost.
> 
> Could you re-check this logic? Most probably I missed something, but still...


  handle_swbp()
    guard(srcu)(&uprobes_srcu);
    handle_chain();
      prepare_uretprobe()
        __srcu_read_lock(&uretprobe_srcu);


vs

  uprobe_free_stage2
    kfree(uprobe)

  uprobe_free_stage1
    call_srcu(&uretprobe_srcu, &uprobe->rcu, uprobe_free_stage2);

  put_uprobe()
    if (refcount_dec_and_test)
      call_srcu(&uprobes_srcu, &uprobe->rcu, uprobe_free_stage1);
  

So my thinking was since we take uretprobe_srcu *inside* uprobe_srcu,
this reference must be visible before we execute stage1, and as such
stage2 cannot complete prematurely.