openstack
/
swift
Code Issues Proposed changes

Merge "wsgi: Start workers in parallel, rather than serially"

This commit is contained in:
Zuul 2022-11-14 17:34:28 +00:00 committed by Gerrit Code Review
parent 61e7ac3304 f08b8e0af2
commit 6a010499d5
1 changed files with 12 additions and 11 deletions
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23
swift/common/wsgi.py
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@ -970,7 +970,7 @@ class ServersPerPortStrategy(StrategyBase):
Called when a new worker is started.
:param socket sock: The listen socket for the worker just started.
:param server_idx: The socket's server_idx as yielded by
:param tuple data: The socket's (port, server_idx) as yielded by
:py:meth:`new_worker_socks`.
:param int pid: The new worker process' PID
"""
@ -1090,6 +1090,7 @@ def run_wsgi(conf_path, app_section, *args, **kwargs):
signal.signal(signal.SIGUSR1, stop_with_signal)
while running_context[0]:
new_workers = {} # pid -> status pipe
for sock, sock_info in strategy.new_worker_socks():
read_fd, write_fd = os.pipe()
pid = os.fork()
@ -1114,16 +1115,16 @@ def run_wsgi(conf_path, app_section, *args, **kwargs):
return 0
else:
os.close(write_fd)
worker_status = os.read(read_fd, 30)
os.close(read_fd)
# TODO: delay this status checking until after we've tried
# to start all workers. But, we currently use the register
# event to know when we've got enough workers :-/
if worker_status == b'ready':
strategy.register_worker_start(sock, sock_info, pid)
else:
raise Exception(
'worker did not start normally: %r' % worker_status)
new_workers[pid] = read_fd
strategy.register_worker_start(sock, sock_info, pid)
for pid, read_fd in new_workers.items():
worker_status = os.read(read_fd, 30)
os.close(read_fd)
if worker_status != b'ready':
raise Exception(
'worker %d did not start normally: %r' %
(pid, worker_status))
# TODO: signal_ready() as soon as we have at least one new worker for
# each port, instead of waiting for all of them