How might `inspect.signature_of_type` work?

(1) classes.py

class Dog:
    def __init__(self, color: str) -> None:
        self.color = color
    def __call__(self, command: str) -> None:
        print("Dog will now " + command)


inspect.signature(Dog("brown")) == inspect.signature(Dog)          # False
inspect.signature(Dog("brown")) == inspect.signature_of_type(Dog)  # True

(2) simple_callables.py

from typing import *

FuncType0 = Callable[[int, str], bool]

class FuncType1(Protocol):
    def __call__(self, __0: int, __1: str, /) -> bool ...

inspect.signature_of_type(FuncType0) == inspect.signature_of_type(FuncType1)

(3) pep612_callables.py

from typing import *

P = ParamSpec("P")

FuncType0a = Callable[P, bool]
FuncType0b = Callable[Concatenate[int, P], bool]

class FuncType1a(Protocol):
    def __call__(self, *__args: P.args, **__kwargs: P.kwargs) -> bool ...

class FuncType1b(Protocol):
    def __call__(self, __0: int, /, *__args: P.args, **__kwargs: P.kwargs) -> bool ...

inspect.signature_of_type(FuncType0a) == inspect.signature_of_type(FuncType1a)
inspect.signature_of_type(FuncType0b) == inspect.signature_of_type(FuncType1b)

(4) pep646_callables.py

from typing import *

Ts_varadic = TypeVarTuple("Ts")


FuncType0a = Callable[[int, *Ts], bool]

class FuncType1a(Protocol, Generic[Ts]):
    def __call__(self, __0: int, *__args: *Ts) -> bool ...

    
inspect.signature_of_type(FuncType0a) == inspect.signature_of_type(FuncType1a)


# But what about this? We can't have a positional-only argument after the vararg,
# so there's no way to express this as a callback protocol!
#
# We could still produce a Signature where the final `str` is marked as
# POSITIONAL_ONLY in spite of coming after the vararg. This probably makes sense,
# but such a signature would be impossible to produce using the existing
# `inspect.signature` function since no method could actually expose that signature.
FuncType0a = Callable[[int, *Ts, str], bool]