为什么不将一个函数声明为type.CoroutineType类型的异步函数? [英] Why isn't a function declared as async of type types.CoroutineType?

问题描述

从此处引用:

types.CoroutineType

协程对象的类型，由异步def函数创建.

The type of coroutine objects, created by async def functions.

此处的引用:

使用async def语法定义的函数始终是协程函数，即使它们不包含await或async关键字.

Functions defined with async def syntax are always coroutine functions, even if they do not contain await or async keywords.

Python控制台会话:

Python console session:

Python 3.5.1 (v3.5.1:37a07cee5969, Dec  6 2015, 01:38:48) [MSC v.1900 32
bit (Intel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import types
>>> def f(): pass
...
>>> async def g(): pass
...
>>> isinstance(f, types.FunctionType)
True
>>> isinstance(g, types.FunctionType)
True
>>> isinstance(g, types.CoroutineType)
False
>>>

为什么isinstance(g, types.CoroutineType)不评估为True?

推荐答案

协程和协程函数之间有区别.生成器和生成器功能之间存在差异的方法相同:

There's a difference between coroutine and coroutine function. The same way as there is a difference between generator and generator function:

调用函数g会返回一个协程，例如:

Calling the function g returns a coroutine, e.g.:

>>> isinstance(g(), types.CoroutineType)
True

如果您需要确定g是否是协程函数(即将返回协程)，则可以使用以下方法进行检查:

If you need to tell if g is a coroutine function (i.e. would return a coroutine) you can check with:

>>> from asyncio import iscoroutinefunction
>>> iscoroutinefunction(g)
True

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