Python Typing: declare return value type based on function argument

You are looking for typing.Type, so something to the effect of:

from typing import TypeVar, Type

T = TypeVar("T", str, complex, float, int)

def fun(t: Type[T]) -> T:
    return t(42)

fun(int)
fun(float)
fun(complex)
fun(str)

Note, your type variable needs to be constrained, because not all Type objects accept arguments, but you can constrain it to a few that do like your example.

TLDR: You need a TypeVar for the return type of calling t:

def fun(t: Callable[[int], R]) -> R:
    ...

---

Constraining on a type is too restrictive here. The function accepts any Callable that takes an integer, and the return type of the function is that of the Callable. This can be specified using a TypeVar for the return type:

from typing import Callable, TypeVar


R = TypeVar('R')  # the variable return type


def fun(t: Callable[[int], R]) -> R:
    return t(42)

fun(int)                            # Revealed type is 'builtins.int*'
fun(float)                          # Revealed type is 'builtins.float*'
reveal_type(fun(lambda x: str(x)))  # Revealed type is 'builtins.str*'

This works for types as well, because type instantiation is a call.

If a more complex signature, e.g. with keyword arguments, is needed, use Protocol (from typing or typing_extensions).

---

Note that if one explicitly wants to pass only 42 to the Callable, Literal (from typing or typing_extensions) can be used to specify that.

R = TypeVar('R')


def fun(t: Callable[[Literal[42]], R]) -> R:
    return t(42)

Note that any function of the type Callable[[int], R] also satisfies Callable[[Literal[42]], R].