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