Question

我前段时间开始学习Tornado框架。我已经面对缺乏经验用户的文档，并检查了asyncio模块文档。所以问题是，我在asyncio中有一些简单的代码：

    import asyncio

    @asyncio.coroutine
    def compute(x, y):
        print("Compute %s + %s ..." % (x, y))
        yield from asyncio.sleep(1.0)
        return x + y


    @asyncio.coroutine
    def print_sum(x, y):
        result = yield from compute(x, y)
        print("%s + %s = %s" % (x, y, result))


loop = asyncio.get_event_loop()
loop.run_until_complete(print_sum(1, 2))
loop.close()

然后我尝试使用Tornado框架做同样的事情：

from tornado.ioloop import IOLoop
from tornado import gen

@gen.coroutine
def compute(x, y):
   print("Compute %s + %s ..." % (x, y))
   yield gen.sleep(1.0)
   return (x+y)

@gen.coroutine
def print_sum(x, y):
   result = yield compute(x, y)
   print("%s + %s = %s" % (x, y, result))

IOLoop.instance().run_sync(print_sum(1,2))

但遗憾的是Tornado代码引发了这样的例外：

 Compute 1 + 2 ...
Traceback (most recent call last):
  File "tornado_coroutine.py", line 19, in <module>
    IOLoop.instance().run_sync(print_sum(1, 2))
  File "C:\Python34\lib\site-packages\tornado\ioloop.py", line 421, in run_sync
    return future_cell[0].result()
  File "C:\Python34\lib\site-packages\tornado\concurrent.py", line 209, in resul
t
    raise_exc_info(self._exc_info)
  File "<string>", line 3, in raise_exc_info
  File "C:\Python34\lib\site-packages\tornado\ioloop.py", line 402, in run
    result = func()
TypeError: 'Future' object is not callable

也许IOLoop试图创造一个新的＆＃34;圈＆＃34;在所有协同程序都返回了它们的值之后？

Answer 1

run_sync将一个函数（或其他＆＃34;可调用的＆＃34;）作为参数。您正在调用该函数，然后将结果作为参数。您只需使用lambda：

即可创建匿名函数

IOLoop.instance().run_sync(lambda: print_sum(1,2))

龙卷风框架。 TypeError：＆＃39;未来＆＃39;对象不可调用

1 个答案: