[FIXED] Python asyncio: Can‘t debug into Task class

Issue

When working with Python's asyncio package, I've noticed that I can't step into any code of its tasks.Task class. For example, when the calling code invokes the class's constructor, my next 'step into' get's me into a get_debug() function outside the class. After that, I return to the calling code with an initialised Task object. I've observed similar behaviour with Task.__next_step(): I'll just step into code that gets called by this method.

All Python versions (3.9, 3.10), IDEs (PyCharm, Visual Studio Code) and OSs (macOS, Windows) that I tested showed the same issue.

Does anyone know the reason for the debugger’s strange behaviour and, possibly, how to overcome it?

The call_soon() in the last line of the screenshot is issued from within Task.__init__. However, as you can see, the debugger never stepped into the initializer.

Update: Surprisingly, with Python 3.6 (Pythonista on iPadOS) I can step into Task.__init__ from base_events.BaseEventLoop.create_task().

Solution

The implementation of Task is in C (where available). The debugger cannot step into C code.

You can see this in the asyncio.tasks module:

class Task(futures._PyFuture):
    ...


_PyTask = Task


try:
    import _asyncio
except ImportError:
    pass
else:
    # _CTask is needed for tests.
    Task = _CTask = _asyncio.Task

That last line shows the Python implementation being overridden by the C implementation.

You can verify which implementation you have by inspecting the __module__ attribute of Task. e.g.

import asyncio
print(asyncio.Task.__module__)

A pure Python implementation will print asyncio.tasks. The C implementation will print _asyncio.

Answered By - Dunes