RUF006 does not catch violations in lambda expressions

[hpelwintervold]

Ruff does not appear to find violations for rule RUF006 (link) in lambda expressions. Example below:

• List of keywords searched before opening:
  RUF006, lambda

• Minimal code snippet:
  sandbox.py

import asyncio


async def foo():
    return

f = lambda *args: asyncio.create_task(foo())

• Command invoked:
  ruff check --select RUF006 --isolated sandbox.py

• Current ruff settings
  N/A - overridden in command

• Ruff version
  ruff 0.6.7

[zanieb]

Thanks for the report. This is a bit tricky, as the return value of f could be stored in a variable later. This is technically not a violation as-written, right? Because create_task is never called?

Implementation-wise, I tried adding this rule to the existing expression visitor but this causes a bunch of false positives when the variable is assigned. I think we need to write a Visitor specifically to find call expressions in lambdas?

I wrote some test cases

f = lambda *args: asyncio.create_task(foo())
f = lambda *args: lambda *args: asyncio.create_task(foo())
f = lambda *args: [asyncio.create_task(foo()) for x in args]

[hpelwintervold]

Thanks for taking a look, the more specific use case I have where I ran into this is signal handling. The example (which does not assign the lambda to anything)

import asyncio
import signal


async def foo():
    return

signal.signal(signal.SIGINT, lambda *args: asyncio.create_task(foo()))

Ruff does not report an error for this which prompted me to raise this bug. But thinking about this some more, I was not understanding lambdas correctly. I had thought the lambda expression lambda *args: asyncio.create_task(foo()) essentially mapped to the following function definition:

def create_task():
    asyncio.create_task(foo())

And ruff emits an error:

sandbox2.py:10:5: RUF006 Store a reference to the return value of `asyncio.create_task`
   |
 9 | def create_task():
10 |     asyncio.create_task(foo())
   |     ^^^^^^^^^^^^^^^^^^^^^^^^^^ RUF006
11 | 
12 | async def sigint_handler(*args):
   |

Found 1 error.

But actually the lambda returns the value - so an equivalent function is this:

def create_task():
    return asyncio.create_task(foo())

Ruff does not emit an error for this, because the reference is returned (not lost), right?

Whether or not that reference is later dropped or not seems to be outside of the scope of RUF006. It's a little tricky to think though, but I think the current behavior of not reporting an error makes sense - because it's up to the caller of the lambda to track the reference to the result. So I think this issue could be closed

[zanieb]

Thanks for following up!