runner: Show kernel state on detecting test timeout

When our watchdog expires and we declare the test has timed out, we send
it a signal to terminate. The test will produce a backtrace upon receipt
of that signal, but often times (especially as we do test and debug the
kernel), the test is hung inside the kernel. So we need the kernel state
to see where the live/deadlock is occuring. Enter sysrq-t to show the
backtraces of all processes (as the one we are searching for may be
sleeping).

Signed-off-by: Chris Wilson <chris@chris-wilson.co.uk>
Cc: Petri Latvala <petri.latvala@intel.com>

1 files changed, 21 insertions, 0 deletions

diff --git a/runner/executor.c b/runner/executor.c
index 1a00237f..2bb82827 100644
--- a/runner/executor.c
+++ b/runner/executor.c
@@ -638,6 +638,25 @@ static const char *get_cmdline(pid_t pid, char *buf, ssize_t len)
 	return buf;
 }
 
+static bool sysrq(char cmd)
+{
+	bool success = false;
+	int fd;
+
+	fd = open("/proc/sysrq-trigger", O_WRONLY);
+	if (fd >= 0) {
+		success = write(fd, &cmd, 1) == 1;
+		close(fd);
+	}
+
+	return success;
+}
+
+static void show_kernel_task_state(void)
+{
+	sysrq('t');
+}
+
 /*
  * Returns:
  *  =0 - Success
@@ -728,6 +747,8 @@ static int monitor_output(pid_t child,
 
 			switch (killed) {
 			case 0:
+				show_kernel_task_state();
+
 				if (settings->log_level >= LOG_LEVEL_NORMAL) {
 					outf("Timeout. Killing the current test with SIGQUIT.\n");
 					fflush(stdout);