Hi,
On Mon, Aug 18, 2025 at 07:33:32AM +0000, Ye Weihua wrote:
> This warning was triggered during testing on v6.16:
>
> notifier callback ftrace_suspend_notifier_call already registered
> WARNING: CPU: 2 PID: 86 at kernel/notifier.c:23
> notifier_chain_register+0x44/0xb0
> ...
> Call Trace:
> <TASK>
> blocking_notifier_chain_register+0x34/0x60
> register_ftrace_graph+0x330/0x410
> ftrace_profile_write+0x1e9/0x340
> vfs_write+0xf8/0x420
> ? filp_flush+0x8a/0xa0
> ? filp_close+0x1f/0x30
> ? do_dup2+0xaf/0x160
> ksys_write+0x65/0xe0
> do_syscall_64+0xa4/0x260
> entry_SYSCALL_64_after_hwframe+0x77/0x7f
>
> When writing to the function_profile_enabled interface, the notifier was
> not unregistered after start_graph_tracing failed, causing a warning the
> next time function_profile_enabled was written.
>
> Fixed by adding unregister_pm_notifier in the exception path.
>
> Fixes: 4a2b8dda3f870 ("tracing/function-graph-tracer: fix a regression while
> suspend to disk")
> Signed-off-by: Ye Weihua <yeweih...@huawei.com>
> ---
> kernel/trace/fgraph.c | 1 +
> 1 file changed, 1 insertion(+)
>
> diff --git a/kernel/trace/fgraph.c b/kernel/trace/fgraph.c
> index c5b207992fb4..dac2d58f3949 100644
> --- a/kernel/trace/fgraph.c
> +++ b/kernel/trace/fgraph.c
> @@ -1391,10 +1391,11 @@ int register_ftrace_graph(struct fgraph_ops *gops)
> error:
> if (ret) {
> ftrace_graph_active--;
> gops->saved_func = NULL;
> fgraph_lru_release_index(i);
> + unregister_pm_notifier(&ftrace_suspend_notifier);
Is this really correct ? The pm notifier is only registered if
ftrace_graph_active==1, but not if it is larger than that.
The above code unregisters it unconditionally, even if
ftrace_graph_active > 1. I can see that the resulting double
unregistration in unregister_ftrace_graph() doesn't really
matter since the error return will be ignored, but is it really
irrelevant for the successful registered graphs no longer get the
benefit of the pm notifier callback ?
Guenter
