Support for `ParamSpec` for `type`

The problem I'm having is that the type doesn't support ParamSpec, so I can't use that to enforce args and kwargs, in the same way as as Callable. I'm not 100% sure whether this is correct, but isn't every type a Callable, so in a sense that type is a subtype of Callable, so there should be a way to represent type in the same way as a Callable to obey Liskov?

For example, I'm trying to write an overload such that passing in a type will return ExpectType, while a general Callable returns ExpectCallable:

from collections.abc import Callable

import typing_extensions as t

P = t.ParamSpec("P")
T_co = t.TypeVar("T_co", covariant=True)

class ExpectCallable(t.Generic[P, T_co]): ...
class ExpectType(ExpectCallable[P, T_co]): ...

@t.overload
def expect(
    v: type[T_co], /, *args: P.args, **kwargs: P.kwargs
) -> ExpectType[P, T_co]: ...
@t.overload
def expect(
    v: Callable[P, T_co], /, *args: P.args, **kwargs: P.kwargs
) -> ExpectCallable[P, T_co]: ...
def expect(
    v: type[T_co] | Callable[P, T_co], /, *args: P.args, **kwargs: P.kwargs
) -> ExpectType[P, T_co] | ExpectCallable[P, T_co]:
    return ExpectType() if isinstance(v, type) else ExpectCallable()

class A:
    def __init__(self, inp: str, /) -> None: ...

def fn(inp: str, /) -> None: ...

t.assert_type(expect(A), ExpectType[[], A])  # Shouldn't be allowed  
t.assert_type(expect(fn, "inp"), ExpectCallable[[str], None])

As you can see that t.assert_type(expect(A), ExpectType[[], A]) passes with no errors, even though A is suppose to take in a str argument. I'm expecting expect(A) to require a str argument, similar to how expect(fn, "inp") requires the second parameter "inp".

Note that I haven't used *args and **kwargs in the example for simplicity, but they are meant to be used inside of the Expect classes.

Maybe the simplest way is to add a new class TypeCallable that allows for TypeCallable[P, T] instead of modifying type, so the above example is achievable:

@t.overload
def expect(
    v: TypeCallable[P, T_co], /, *args: P.args, **kwargs: P.kwargs
) -> ExpectType[P, T_co]: ...
@t.overload
def expect(
    v: Callable[P, T_co], /, *args: P.args, **kwargs: P.kwargs
) -> ExpectCallable[P, T_co]: ...
def expect(
    v: TypeCallable[P, T_co] | Callable[P, T_co], /, *args: P.args, **kwargs: P.kwargs
) -> ExpectType[P, T_co] | ExpectCallable[P, T_co]:
    return ExpectType() if isinstance(v, type) else ExpectCallable()

Note that I think https://github.com/python/typing/issues/1966 might also solve the above problem.

查看示例中描述的 ParamSpec 对 type 和 Callable 的行为，然后将其与 typing issue #1966 中的可能方向进行比较。完成这项工作需要达成一种一致的表示方式，该表示方式能够保留构造函数参数信息，并使所提议的重载可区分，但此 issue 没有指明实现文件或测试。