zmq send with NOBLOCK raise Resource temporarily unavailable
问题 This code will raise Resource temporarily unavailable when call with NOBLOCK: context = zmq.Context() sender = context.socket(zmq.PUSH) sender.bind('tcp://*:15556') sender.send('KeEpAliv', zmq.NOBLOCK) # this line will throw exception #sender.send('KeEpAliv') # this line will ok After read the docs, I found no hints for this. but docs for recv explained this flag. 回答1: Python wrappers raise zmq.error.Again if the underlying C API returns EAGAIN . Now, you should follow to zmq_send