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(馃巵) Perform 'union math' on non overloaded calls聽#13493
Description
Here the call to f is distributing the generic over the return type (based), but the call to the standalone function can't be resolved
from typing import Iterable, overload, TypeVar
a: list[int] | list[str]
@overload
def f[T](i: list[T]) -> list[T]: ...
@overload
def f() -> None: ...
def f[T](i: list[T] = []) -> list[T] | None: ...
reveal_type(f(a)) # list[int] | list[str]
def g[T](i: list[T]) -> list[T]: return []
reveal_type(g(a)) # error: Cannot infer type argument 1 of "g" [misc]What's happening here is that union_overload_result, which is intended for when a call with a union would match multiple overload parts, is matching twice against the same overload, resulting is a more accurate return type.
Activity
[-](馃悶) Generics behave differently on overloads compared to single functions[/-][+](馃巵) Perform 'union math' on non overloaded calls[/+]ilevkivskyi commentedon Feb 9, 2025
Both examples now infer this:
But this is by accident, because in your example
T = int | stris a valid solution forTeven without union math. But in general I agree that we may want to use union math more eagerly if it results in a more precise type.KotlinIsland commentedon Feb 19, 2025
i updated the op