qa/vstart_runner: don't let command run after timeout

Parameter "timeout" is accepted by LocalRemote.run() but the method
doesn't do anything about it besides accepting it. Thus, this parameter
has no effect.

In LocalRemote.run(), pass this parameter to LocalRemoteProcess.wait()
and from this method pass it to subprocess.Popen.communicate(). Thus,
command will be terminated by subprocess module at seconds specified by
"timeout" parameter. IOW, "timeout" parameter will have an effect.

Fixes: https://tracker.ceph.com/issues/65533
Signed-off-by: Rishabh Dave <ridave@redhat.com>

1 files changed, 3 insertions, 3 deletions

diff --git a/qa/tasks/vstart_runner.py b/qa/tasks/vstart_runner.py
index 8d9dc6f1845..b5dbe3c5f97 100644
--- a/qa/tasks/vstart_runner.py
+++ b/qa/tasks/vstart_runner.py
@@ -233,7 +233,7 @@ class LocalRemoteProcess(object):
         else:
             self.stderr.write(err)
 
-    def wait(self):
+    def wait(self, timeout=None):
         # Null subproc.stdin so communicate() does not try flushing/closing it
         # again.
         if self.stdin is not None and self.stdin.closed:
@@ -249,7 +249,7 @@ class LocalRemoteProcess(object):
             else:
                 return
 
-        out, err = self.subproc.communicate()
+        out, err = self.subproc.communicate(timeout=timeout)
         out, err = rm_nonascii_chars(out), rm_nonascii_chars(err)
         self._write_stdout(out)
         self._write_stderr(err)
@@ -488,7 +488,7 @@ sudo() {
         )
 
         if wait:
-            proc.wait()
+            proc.wait(timeout)
 
         return proc