Cannot infer type of lambda when using default arguments

Bug Report

I get error: Cannot infer type of lambda even though the code seems fine to me.

Currently, I don't see any other workaround other than using # type: ignore.

To Reproduce

from typing import Callable, List

def foo(factories: List[Callable[[], int]]):
    return sum(f() for f in factories)

names = ["0", "abc"]
# Inference works fine
assert foo([lambda: len(name) for name in names]) == 4
# Cannot infer type of lambda
assert foo([lambda name=name: len(name) for name in names]) == 4

def foo2(factories: List[Callable[[int], int]]):
    return sum(f(0) for f in factories)

# Inference works fine
assert foo2([lambda i: len(name) for name in names]) == 4
# Cannot infer type of lambda
assert foo2([lambda i, name=name: len(name) for name in names]) == 4

Checked with:

mypy mypy-lambda-test.py

Expected Behavior

There should be no error. Especially because the other variants without default arguments are deemed fine.

Furthermore, using default arguments like this is the standard workaround to capture loop variables to lambdas.

Your Environment

• Mypy version used: 0.942
• Python version used: 3.9.7
• Operating system and version: Ubuntu 21.10

Start with the lambda examples in mypy-lambda-test.py and run mypy mypy-lambda-test.py using the reported Python and mypy versions. Compare the working lambdas with the variants that capture name through a default argument. Done means all four assertions type-check without requiring # type: ignore.