tracing: Do note expose stack_trace_filter without DYNAMIC_FTRACE

The "stack_trace_filter" file only makes sense if DYNAMIC_FTRACE is
configured in. If it is not, then the user can not filter any functions.

Not only that, the open function causes warnings when DYNAMIC_FTRACE is not
set.

Link: http://lkml.kernel.org/r/20170710110521.600806-1-arnd@arndb.de

Reported-by: Arnd Bergmann <arnd@arndb.de>
Signed-off-by: Steven Rostedt (VMware) <rostedt@goodmis.org>
This commit is contained in:
Steven Rostedt (VMware) 2017-07-11 19:21:04 -04:00
parent dc8d387210
commit bbd1d27d86

View File

@ -406,6 +406,8 @@ static const struct file_operations stack_trace_fops = {
.release = seq_release,
};
#ifdef CONFIG_DYNAMIC_FTRACE
static int
stack_trace_filter_open(struct inode *inode, struct file *file)
{
@ -423,6 +425,8 @@ static const struct file_operations stack_trace_filter_fops = {
.release = ftrace_regex_release,
};
#endif /* CONFIG_DYNAMIC_FTRACE */
int
stack_trace_sysctl(struct ctl_table *table, int write,
void __user *buffer, size_t *lenp,
@ -477,8 +481,10 @@ static __init int stack_trace_init(void)
trace_create_file("stack_trace", 0444, d_tracer,
NULL, &stack_trace_fops);
#ifdef CONFIG_DYNAMIC_FTRACE
trace_create_file("stack_trace_filter", 0444, d_tracer,
&trace_ops, &stack_trace_filter_fops);
#endif
if (stack_trace_filter_buf[0])
ftrace_set_early_filter(&trace_ops, stack_trace_filter_buf, 1);