CODEWITHSUNDEEP
pythonpython-3.5python-asyncio
Szabolcs Dombi
Szabolcs Dombi

Reputation: 5783

Program hang with asyncio

Here is my code:

import asyncio, socket

sock = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
sock.bind(('', 1234))
sock.setblocking(False)

queue = asyncio.Queue()

def sock_reader():
    print(sock.recv(1024))
    # x = yield from queue

def test_sock_reader():
    sock = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
    sock.sendto(b'HELLO', ('127.0.0.1', 1234))

loop = asyncio.get_event_loop()
loop.add_reader(sock, sock_reader)
loop.call_later(0.5, test_sock_reader)
loop.run_forever()
loop.close()

This is the output:

b'HELLO'

When the line # x = yield from queue is uncommented the program is not printing b'Hello' anymore.

Why is the yield from affecting a command that should already be executed?

Upvotes: 2

Views: 592

Answers (1)

bagrat
bagrat

Reputation: 7418

The problem is a combination of syntax and API definition.

First of, refer to the documentation of add_reader, which states that it expects a callback. It is not obvious from the word itself, but by saying callback it means a regular function.

Now, when you uncomment the # x = yield from queue line, your sock_reader function actually becomes a generator/coroutine due to yield from, in which case when called like a regular function (i.e. sock_reader(...)), it returns a generator object, and does not get executed.

Upvotes: 5

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