From 54008ad24b8883a101058de93cd9cdfe209db994 Mon Sep 17 00:00:00 2001 From: Golo Roden Date: Sat, 24 Oct 2026 10:23:00 +0200 Subject: [PATCH] chore: Leave the default startup timeout to the test of a container that exits, so that a busy Docker does not run out of time before the container prints its log. The test gave every attempt 5 seconds, which a busy Docker used up while creating and starting the container, so that the start failed with "Docker did not answer within 5s" instead of the log tail. The test does not depend on the timeout, since a start that waited for the time to run out would fail with another error. Co-Authored-By: Claude Opus 5.5 Claude-Session: https://claude.ai/code/session_01Svp844n6XkEuoe7y6zEFtF --- tests/test_container.py | 6 +++++- 1 file changed, 5 insertions(+), 1 deletion(-) diff --git a/tests/test_container.py b/tests/test_container.py index 05c83ca..b274637 100644 --- a/tests/test_container.py +++ b/tests/test_container.py @@ -458,7 +458,11 @@ def test_fails_at_once_with_the_log_tail_if_the_container_exits( 'for line in $(seq 1 30); do echo "line $line"; done; exit 1', ], } - container = Container().with_image_tag(get_image_tag()).with_startup_timeout(5) + # The default startup timeout leaves a busy Docker the time it needs to + # create and start the container. The test does not depend on it: a + # start that waited for the time to run out would fail with another + # error than the log tail. + container = Container().with_image_tag(get_image_tag()) with pytest.raises(RuntimeError) as error: container.start()