我们如何调用需要协程的普通函数? [英] How do we call a normal function where a coroutine is expected?

问题描述

考虑一个调用另一个协程的协程:

Consider a coroutine which calls into another coroutine:

async def foo(bar):
     result = await bar()
     return result

如果 bar 是一个协程，这可以正常工作.如果 bar 是一个普通函数，我需要做什么(即我需要做什么来包装对 bar 的调用)，以便此代码执行正确的操作?

This works fine if bar is a coroutine. What do I need to do (i.e. with what do I need to wrap the call to bar) so that this code does the right thing if bar is a normal function?

使用 async def 定义协程是完全可能的，即使它从不做任何异步操作(即从不使用 await).但是，问题询问如何在 foo 的代码中包装/修改/调用常规函数 bar，以便可以等待 bar.

It is perfectly possible to define a coroutine with async def even if it never does anything asynchronous (i.e. never uses await). However, the question asks how to wrap/modify/call a regular function bar inside the code for foo such that bar can be awaited.

推荐答案

只需使用 asyncio.coroutine 如果需要:

Simply wrap your synchronous function with asyncio.coroutine if needed:

if not asyncio.iscoroutinefunction(bar):
    bar = asyncio.coroutine(bar)

由于重新包装协程是安全的，所以实际上不需要协程功能测试:

Since it is safe to re-wrap a coroutine, the coroutine function test is actually not required:

async_bar = asyncio.coroutine(sync_or_async_bar)

因此，您的代码可以重写如下:

Therefore, your code can be re-written as follows:

async def foo(bar):
     return await asyncio.coroutine(bar)()

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