Unknown `DefaultNamedArg` raises type incompatibility errors

🐛 Bug Report

DefaultNamedArg may cause signature compatibility errors. Observed when a function has a default keyword argument in signature and is passed to be used as a function without that particular keyword argument.

To Reproduce

from typing import List, AnyStr
from glob import glob
import itertools as it
def expand_patterns(
        pats: List[AnyStr],
        sort=True
):
    paths = it.chain(map(glob, pats))
    # Argument 1 to "map" has incompatible type "Callable[[AnyStr, DefaultNamedArg(bool, 'recursive')], List[AnyStr]]"; expected "Callable[[AnyStr], List[AnyStr]]"
    if sort:
        paths = sorted(paths)
    return paths

Expected Behavior

Not sure if this ought to happen, but I'd expect a function like foo(x: int, y='y') to be peacefully passable to a function expecting Callable[[int], Any].

Actual Behavior

# Argument 1 to "map" has incompatible type "Callable[[AnyStr, DefaultNamedArg(bool, 'recursive')], List[AnyStr]]"; expected "Callable[[AnyStr], List[AnyStr]]"

Your Environment

• Mypy version used: 0.782
• Mypy command-line flags: Default of VSCode
• Mypy configuration options from mypy.ini (and other config files): N/A
• Python version used: 3.7.7
• Operating system and version: Windows 10, 2004

Commencez par reproduire l’exemple Python fourni qui fait intervenir glob, map, la compatibilité de Callable et DefaultNamedArg. Suivez la manière dont mypy compare les signatures de Callable, puis vérifiez qu’un argument nommé par défaut est traité conformément à la règle de compatibilité prévue et que l’incompatibilité signalée est résolue.