Narrowed down types sometimes not propagated to lambda

[JukkaL]

The errors generated for the following code are invalid (when using --strict-optional):

from typing import Optional, Callable

def f1(key: Callable[[], str]) -> None: ...
def f2(key: object) -> None: ...

def g(b: Optional[str]) -> None:
    if b:
        f1(lambda: b.upper())  # Item "None" of "Optional[str]" has no attribute "upper"
        z: Callable[[], str] = lambda: b.upper()  # Item "None" of "Optional[str]" has no attribute "upper"
        f2(lambda: b.upper())  # No error
        lambda: b.upper()  # No error

My hypothesis is that callable type context for lambda somehow affects how type narrowing behaves.

This was reported by @tueda on Gitter.

[ikonst]

Workaround:

def g(b: Optional[str]) -> None:
    if b:
        _b = b  # https://github.com/python/mypy/issues/4297
        f1(lambda: _b.upper())

[calimeroteknik]

This also seems to happen with native collection types whose size is "runtime" (i.e. recursive types in lambda-calculus terms), but not tuples (whose size is "fixed").

Error / non-error cases added for reproduction, based on the original post:

from typing import Optional, Callable

def f1(key: Callable[[], str]) -> None: ...
def f2(key: object) -> None: ...

def g(b: Optional[str]) -> None:
    if b is not None:
        f1(lambda: b.upper())  # Item "None" of "Optional[str]" has no attribute "upper"
        z: Callable[[], str] = lambda: b.upper()  # Item "None" of "Optional[str]" has no attribute "upper"
        d = { "foo": lambda: b.upper() }  # Item "None" of "Optional[str]" has no attribute "upper"
        l = [ lambda: b.upper() ]  # Item "None" of "Optional[str]" has no attribute "upper"
        t = ( lambda: b.upper(), ) # No error
        f2(lambda: b.upper())  # No error
        lambda: b.upper()  # No error