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

Inizia con gli esempi lambda in mypy-lambda-test.py ed esegui mypy mypy-lambda-test.py usando le versioni indicate di Python e mypy. Confronta le lambda funzionanti con le varianti che catturano name tramite un argomento predefinito. Il lavoro è completato quando tutte e quattro le asserzioni superano il controllo dei tipi senza richiedere # type: ignore.