reveal_type() may affect inferred type / overload resolution

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.

Comienza con el reproductor de Python proporcionado usando --py2, comparando filter(None, reveal_type(a)) con filter(None, a) y el error de asignación informado. Rastrea cómo reveal_type afecta a los tipos inferidos y a la resolución de sobrecargas; se considera terminado cuando ambas llamadas producen el mismo resultado inferido y reveal_type sigue siendo únicamente diagnóstico.