Update comment about PID reuse race

Author: giampaolo

Lib/subprocess.py

@@ -2125,20 +2125,18 @@ def _wait(self, timeout):
                 # Try efficient wait first.
                 if self._wait_pidfd(timeout) or self._wait_kqueue(timeout):
                     # Process is gone. At this point os.waitpid(pid, 0)
-                    # should return immediately, but in rare races
-                    # another thread or signal handler may have already
-                    # reaped the PID. os.waitpid(pid, WNOHANG) ensures
-                    # we attempt a non-blocking reap safely without
-                    # blocking indefinitely.
+                    # should return immediately, but in rare races the
+                    # PID may have been reused.
+                    # os.waitpid(pid, WNOHANG) ensures we attempt a
+                    # non-blocking reap without blocking indefinitely.
                     with self._waitpid_lock:
                         if self.returncode is not None:
                             return self.returncode
                         (pid, sts) = self._try_wait(os.WNOHANG)
                         assert pid == self.pid or pid == 0
                         if pid == self.pid:
                             self._handle_exitstatus(sts)
-                            return self.returncode
-                        return None
+                        return self.returncode
 
                 # Enter a busy loop if we have a timeout.  This busy loop was
                 # cribbed from Lib/threading.py in Thread.wait() at r71065.