Python Typing: declare return value type based on function argument

Question

Suppose I have function that takes type as argument and returns instance of that type:

def fun(t):
    return t(42)

Then I can call it and get objects of provided types:

fun(int)           # 42
fun(float)         # 42.0
fun(complex)       # (42+0j)
fun(str)           # "42"
fun(MyCustomType)  # something

_{That list is not exhaustive, I'd like to be able to use any type with appropriate constructor.}

Then, I'd like to add type hints for that function. What should be the type hint for return value of that function?

---

I've tried using simply t, as t is a type:

def fun(t: type) -> t:
    return t(42)

but that doesn't work:

main.py:1: error: Name 't' is not defined

---

This answer suggests using a TypeVar:

from typing import TypeVar

T = TypeVar("T")

def fun(t: T) -> T:
    return t(42)

But that doesn't seem to be right, as T denotes a type, so it suggests that type itself is returned, not its instance. Mypy rejects it:

main.py:6: error: "object" not callable

---

Using Any obviously work, but I feel it's too vague, it doesn't convey the intent:

from typing import Any

def fun(t: type) -> Any:
    return t(42)

Answer 1

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].

Answer 2

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.