Skip to content

bugfix: retry exponential backoff max float overflow #48378

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Closed
alealandreev wants to merge 2 commits into from
Closed

bugfix: retry exponential backoff max float overflow #48378

Changes from all commits
Commits
Show all changes
2 commits
Select commit Hold shift + click to select a range
84ba360
Issue47971SolutionPatch
alealandreev Mar 26, 2025
5c640de
Merge branch 'main' into patch-3
alealandreev Apr 3, 2025
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
6 changes: 5 additions & 1 deletion airflow-core/src/airflow/models/taskinstance.py
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -2379,14 +2379,18 @@ def next_retry_datetime(self):
"""
from airflow.models.abstractoperator import MAX_RETRY_DELAY

max_try_number = 500
delay = self.task.retry_delay
if self.task.retry_exponential_backoff:
# If the min_backoff calculation is below 1, it will be converted to 0 via int. Thus,
# we must round up prior to converting to an int, otherwise a divide by zero error
# will occur in the modded_hash calculation.
# this probably gives unexpected results if a task instance has previously been cleared,
# because try_number can increase without bound
min_backoff = math.ceil(delay.total_seconds() * (2 ** (self.try_number - 1)))
if self.try_number < max_try_number:
min_backoff = math.ceil(delay.total_seconds() * (2 ** (self.try_number - 1)))
else:
min_backoff = math.ceil(delay.total_seconds() * (2 ** (max_try_number - 1)))

# In the case when delay.total_seconds() is 0, min_backoff will not be rounded up to 1.
# To address this, we impose a lower bound of 1 on min_backoff. This effectively makes
Expand Down