让我说我有一些事件监听器应该一直运行,并且有一些exceptions
我想将它们传递给函数调用者之类的东西
import asyncio
_ = 0
async def listener(loop):
while True:
await asyncio.sleep(0.5)
if _ != 0:
raise ValueError('the _ is not 0 anymore!')
print('okay')
async def executor(loop):
while True:
x = await loop.run_in_executor(None, input, 'execute: ')
global _
_ = x
async def main(loop):
asyncio.ensure_future(listener(loop), loop=loop)
await executor(loop)
loop = asyncio.get_event_loop()
loop.run_until_complete(main(loop))
如果您要更改该值,则监听器事件循环将停止,但我不希望它发生break
,我希望它发生raise
错误,这样您就可以捕获它并且loop
继续前进
答案 0 :(得分:0)
我不希望它中断,我希望它引发错误,这样您就可以抓住它并让循环继续下去
如果要引发错误但继续进行,则侦听器不应直接raise
。相反,它应该使用Future
对象向感兴趣的参与者发出异常信号。 main()
不应等待执行者,而应循环等待广播的未来:
import asyncio
_ = 0
broadcast = None
async def listener():
while True:
await asyncio.sleep(0.5)
if _ != 0:
broadcast.set_exception(ValueError('the _ is not 0 anymore!'))
else:
print('okay')
async def executor():
loop = asyncio.get_event_loop()
while True:
x = int(await loop.run_in_executor(None, input, 'execute: '))
global _
_ = x
async def main():
global broadcast
loop = asyncio.get_event_loop()
loop.create_task(listener())
loop.create_task(executor())
while True:
broadcast = loop.create_future()
try:
await broadcast
except ValueError as e:
print('got error', e)
asyncio.get_event_loop().run_until_complete(main())