Question

在我的项目下，我需要重复构造和销毁上下文，但这会导致错误。

例如：

import zmq

for i in range(100):
    print(i)
    context = zmq.Context()
    data_socket = context.socket(zmq.SUB)
    data_socket.connect("tcp://127.0.0.1:5552")
    data_socket.setsockopt_string(zmq.SUBSCRIBE, "")
    context.destroy()

它返回

0
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13
Traceback (most recent call last):
  File "test.py", line 7, in <module>
    data_socket.connect("tcp://127.0.0.1:5552")
  File "zmq/backend/cython/socket.pyx", line 580, in zmq.backend.cython.socket.Socket.connect
  File "zmq/backend/cython/checkrc.pxd", line 25, in zmq.backend.cython.checkrc._check_rc
zmq.error.ZMQError: Socket operation on non-socket

Answer 1

套接字选项必须放在.connect()或.bind()方法之前，并且您可以从zmq.Context()创建一个单元实例。

尝试一下：

import zmq

context = zmq.Context.instance()

for i in range(100):
    print(i)
    data_socket = context.socket(zmq.SUB)
    data_socket.setsockopt(zmq.SUBSCRIBE, b"")
    data_socket.connect("tcp://127.0.0.1:5552")

context.destroy()

[您的答案]：

但是，如果要使用自己的方式，则应在每次迭代中关闭套接字，这样您的代码段将为：

import zmq

for i in range(100):
    ctx = zmq.Context.instance()
    sock = ctx.socket(zmq.SUB)
    sock.setsockopt(zmq.SUBSCRIBE, b'')
    sock.connect('tcp://127.0.0.1:5552')
    sock.close()  # Note
    ctx.destroy()
    print('ctx closed status: ', ctx.closed, ' iteration: ', i)

出局：

('ctx closed status: ', True, ' iteration: ', 0)
('ctx closed status: ', True, ' iteration: ', 1)
('ctx closed status: ', True, ' iteration: ', 2)
('ctx closed status: ', True, ' iteration: ', 3)
('ctx closed status: ', True, ' iteration: ', 4)
('ctx closed status: ', True, ' iteration: ', 5)
('ctx closed status: ', True, ' iteration: ', 6)
('ctx closed status: ', True, ' iteration: ', 7)
('ctx closed status: ', True, ' iteration: ', 8)
('ctx closed status: ', True, ' iteration: ', 9)
('ctx closed status: ', True, ' iteration: ', 10)
('ctx closed status: ', True, ' iteration: ', 11)
('ctx closed status: ', True, ' iteration: ', 12)
('ctx closed status: ', True, ' iteration: ', 13)
('ctx closed status: ', True, ' iteration: ', 14)
.
.
.

反复构造和销毁上下文会给出错误＃1260

1 个答案: