Python

I thought about it, but it isn't as expressive as I wished.

Meaning if I do

a = foo(return_more=True)
or
a, b = foo(return_more=False)

it doesn't catch these errors for me.

In comparison, the other suggested solution does catch these.

Nice! It looks like the best solution out there.

yea, this is pretty close to what I'm looking for.

The only missing piece is the ability to define the overload methods on the bool

something like

@overload
def foo(return_more: True) -> (Data, Data)

@overload
def foo(return_more: False) -> Data

But I don't think such constructs are possible? I know it is possible in Typescript to define the types using constants, but I don't suppose Python allows for this?

EDIT: At first, when I tried the above, the typechecker said Literal[True] was not expected and I thought it was not possible. But after experimenting some, I figured out that it is actually possible. Added my solution to the OP

Thanks for the tip!

This is the real answer, overloads are meant for exactly this purpose.

It’ll be something like this:

from typing import Literal, overload

@overload
def foo() -> Data: …
@overload
def foo(return_more: Literal[True]) -> tuple[Data, Data]: …
def foo(return_more: bool = False) -> Data | tuple[Data, Data]
    ...
    if return_more:
        return data, more_data
   return data

Python >= 3.10 version:

def foo(return_more: bool) -> DataType | tuple[DataType, MoreDataType]: ...

But i would definitely avoid to do that if possible. I would maybe do something like this instead:

def foo(return_more: bool) -> tuple[DataType, MoreDataType | None]:
    ...
    if return_more:
        return data, more_data
   return data, None

Or if data is a dict, just update it with more_data:

def foo(return_more: bool) -> dict[str, Any]:
    ...
    if return_more:
        return data.update(more_data)
   return data

You can also consider the new union that was introduced with Python 3.10, check PEP604 for details:

def foo(return_more: bool) -> Type1 | tuple[Type2,Type3]: