I don't find this example particularly compelling. `TypeGaurd` is not an instantiable type. A function that returns `TypeGaurd[T]` in actuality returns `bool`: """ Return statements within a type guard function should return bool values, and type checkers should verify that all return paths return a bool. """ So this issue of overloading seems pretty irrelevant. On Wed, Mar 17, 2021 at 4:31 PM Jake Bailey via Typing-sig < typing-sig@python.org> wrote:
I found this confusing too, until I looked further into the overload rules and overlapping with classes. Consider this reduced example:
``` from typing import Union, overload
class bool: ... class TypeGuard(bool): ... # Illegal, but pretend it is okay.
@overload def func(x: TypeGuard) -> int: ... @overload def func(x: bool) -> str: ... def func(x: Union[TypeGuard, bool]) -> Union[int, str]: if isinstance(x, TypeGuard): return 1234 return "hey" ```
If I then write:
``` def do_something() -> bool: return TypeGuard()
b: bool = do_something() x = func(b) ```
If we say "it's a `bool`" and attempt to pick the second overload, it will be wrong, because at runtime it's `TypeGuard`, and `func` returns an `int`, not the expected `str`.
See also: https://github.com/python/typing/issues/253#issuecomment-389262904 _______________________________________________ Typing-sig mailing list -- typing-sig@python.org To unsubscribe send an email to typing-sig-leave@python.org https://mail.python.org/mailman3/lists/typing-sig.python.org/ Member address: donovick@cs.stanford.edu