On Tue, 22 Nov 2022 22:32:58 -0500 Steven Rostedt <rostedt@xxxxxxxxxxx> wrote: > On Wed, 23 Nov 2022 11:18:06 +0800 > Zheng Yejian <zhengyejian1@xxxxxxxxxx> wrote: > > > But Yujie Liu <yujie.liu@xxxxxxxxx> reported a problem that highly > > reproducible after applying this patch: > > Link: https://lore.kernel.org/lkml/54f23c9c-97ae-e326-5873-bfa5d2c81f52@xxxxxxxxx/ > > > > So please DO NOT apply this patch before I find what happened about it. > > I know what the issue is. > > The current way of assigning types is to always increment. And not to > reuse until it fills up. And even then, it looks for the next available > number. > > I'm guessing the IDA will reuse a number as soon as it is freed. This Yes, it is. > may also have uncovered a bug, as in reality, we must actually clear > the tracing buffers every time a number is reused. Thanks for your explanation! It seems almost the case, and with current way of assigning types, this problem maybe also happend when reuse type id, am I right? But instead of clear tracing buffers, would it be better to just mark that record invalid if we had a way of knowing that the format had changed? > > What happens is that the type number is associated to a print format. > That is, the raw data is tagged with the type. This type maps to how to > parse the raw data. If you have a kprobe, it creates a new type number. > If you free it, and create another one. With the IDA, it is likely to > reassign the previously freed number to a new probe. > > To explain this better, let's look at the following scenario: > > echo 'p:foo val=$arg1:u64' > kprobe_events > echo 1 > events/kprobes/foo/enable > sleep 1 > echo 0 > events/kprobes/foo/enable > > echo 'p:bar val=+0($arg1):string' > kprobe_events > > # foo kprobe is deleted and bar is created and > # with IDA, bar has the same number for type as foo > > cat trace > > When you read the trace, it will see a binary blob representing an > event and marked with a type. Although the event was foo, it will now > map it to bar. And it will read foo's $arg1:u64 as bar's > +0($arg1):string, and will crash. > > -- Steve
