Python traceback for coroutine

Question

Let's say I have the following code: from types import coroutine @coroutine def stop(): yield 1 async def test2(): await stop() async def test1(): await test2() await test2() # Here await test2() coro = test1() coro.send(None) coro.send(None) How can I get traceback (traceback object) for current coroutine state, i.e. line marked with comment, without artificially throwing unneeded exception? JavaScript questions and answers, JavaScript questions pdf, JavaScript question bank, JavaScript questions and answers pdf, mcq on JavaScript pdf, JavaScript questions and solutions, JavaScript mcq Test , Interview JavaScript questions, JavaScript Questions for Interview, JavaScript MCQ (Multiple Choice Questions)

Answer 1

Use of traceback.print_stack() will give you the exact same traceback as thrown exception: async def test1(): await test2() traceback.print_stack() # raise Exception() await test2() # Here await test2() You can use traceback.extract_stack() if you want to recieve object instead of printing. Note however that you're doing something strange. asyncio coroutines is not supposed to be run use it's generator's nature functions like .send(). In asyncio you await for coroutines and run top level coroutine using event loop. Please see how it's done in documentation. I write another small example that shows how to print start inside inner coroutine when you use asyncio regular way: import asyncio import traceback async def test3(): traceback.print_stack() async def test2(): await test3() async def test1(): await test2() asyncio.run(test1()) You'll see: File "C:\main.py", line 24, in asyncio.run(test1()) # inner event loop stack here File "C:\main.py", line 21, in test1 await test2() File "C:\main.py", line 17, in test2 await test3() File "C:\main.py", line 13, in test3 traceback.print_stack()