progress: always show done message

The done message was omitted if the task is shorter than 0.5s. This
might confuse users.

Bug: b/371638995
Change-Id: I3fdd2cd8daea16d34fba88457d09397fff71af15
Reviewed-on: https://gerrit-review.googlesource.com/c/git-repo/+/440222
Tested-by: Kuang-che Wu <[email protected]>
Commit-Queue: Kuang-che Wu <[email protected]>
Reviewed-by: Josip Sokcevic <[email protected]>

Author: kcwu

progress.py

@@ -100,6 +100,7 @@ def __init__(
         self._show = not delay
         self._units = units
         self._elide = elide and _TTY
+        self._quiet = quiet
 
         # Only show the active jobs section if we run more than one in parallel.
         self._show_jobs = False
@@ -114,13 +115,7 @@ def __init__(
         )
         self._update_thread.daemon = True
 
-        # When quiet, never show any output.  It's a bit hacky, but reusing the
-        # existing logic that delays initial output keeps the rest of the class
-        # clean.  Basically we set the start time to years in the future.
-        if quiet:
-            self._show = False
-            self._start += 2**32
-        elif show_elapsed:
+        if not quiet and show_elapsed:
             self._update_thread.start()
 
     def _update_loop(self):
@@ -160,7 +155,7 @@ def update(self, inc=1, msg=None):
             msg = self._last_msg
         self._last_msg = msg
 
-        if not _TTY or IsTraceToStderr():
+        if not _TTY or IsTraceToStderr() or self._quiet:
             return
 
         elapsed_sec = time.time() - self._start
@@ -202,7 +197,7 @@ def update(self, inc=1, msg=None):
 
     def end(self):
         self._update_event.set()
-        if not _TTY or IsTraceToStderr() or not self._show:
+        if not _TTY or IsTraceToStderr() or self._quiet:
             return
 
         duration = duration_str(time.time() - self._start)