openstack的多进程wsgi服务用了如下方式
def start(self):
"""Start serving a WSGI application.
:returns: None
"""
self.dup_socket = self.socket.dup()
wsgi_kwargs = {
'func': eventlet.wsgi.server,
'sock': self.dup_socket,
'site': self.app,
'protocol': self._protocol,
'custom_pool': self._pool,
# 'log': self._logger,
'log': LOG,
'log_format': self.conf.wsgi_log_format,
'debug': False,
'keepalive': self.conf.wsgi_keep_alive,
'socket_timeout': self.client_socket_timeout
}
if self._max_url_len:
wsgi_kwargs['url_length_limit'] = self._max_url_len
self._server = eventlet.spawn_n(**wsgi_kwargs)
self.socket是init的时候就bind listen好的,start里的代码是子进程里执行的
比较好奇的是为什子进程要用dup的socket而不直接用self.socket
英文的说明如下
# The server socket object will be closed after server exits,
# but the underlying file descriptor will remain open, and will
# give bad file descriptor error. So duplicating the socket object,
# to keep file descriptor usable.
意思是主进程退出会导致socket关闭？
按照道理只要还有进程的fd指向文件表...文件表就不会被关闭
其他进程的fd和文件表照样打开的呀
关闭sock的地方在eventlet.wsgi.server中
finally:
pool.waitall()
serv.log.info("(%s) wsgi exited, is_accepting=%s" % (
serv.pid, is_accepting))
try:
# NOTE: It's not clear whether we want this to leave the
# socket open or close it. Use cases like Spawning want
# the underlying fd to remain open, but if we're going
# that far we might as well not bother closing sock at
# all.
sock.close()
except socket.error as e:
if support.get_errno(e) not in BROKEN_SOCK:
traceback.print_exc()
我估计问题就出在这里
如果没有复制的socket的fd,当前进程内又显示调用了socket.close
则会会触发监听关闭,导致其他进程虽然fd和文件表都在,但是监听已经被关闭了