reveal_type() may affect inferred type / overload resolution

[JukkaL]

The first filter call below generates an error when using --py2, but the same call without reveal_type(...) is fine:

from typing import Any, List

a = 0  # type: Any

x = filter(None, reveal_type(a))  # type: List[str]  # Error
x2 = filter(None, a)  # type: List[str]  # No error

The error message is Incompatible types in assignment (expression has type "Tuple[<nothing>, ...]", variable has type "List[str]").

reveal_type should have no effect on inferred types, since it's often used to diagnose issues with type inference.

[AlexWaygood]

This error is still reproduceable if you run mypy with --python-version 2.7. It's also reproduceable with Python 3 set as the Python version, if you copy-paste typeshed's Python-2 stub for filter into a file:

from typing import *

_T = TypeVar("_T")

a = 0  # type: Any

@overload  # type: ignore[no-overload-impl]
def py2_filter(__function: Callable[[AnyStr], Any], __iterable: AnyStr) -> AnyStr: ...  # type: ignore
@overload
def py2_filter(__function: None, __iterable: tuple[_T | None, ...]) -> tuple[_T, ...]: ...  # type: ignore
@overload
def py2_filter(__function: Callable[[_T], Any], __iterable: tuple[_T, ...]) -> tuple[_T, ...]: ...  # type: ignore
@overload
def py2_filter(__function: None, __iterable: Iterable[_T | None]) -> list[_T]: ...
@overload
def py2_filter(__function: Callable[[_T], Any], __iterable: Iterable[_T]) -> list[_T]: ...

y = py2_filter(None, reveal_type(a))  # type: List[str]  # Error
y2 = py2_filter(None, a)  # type: List[str]  # No error