Python: How to write typing.overload decorator for bool arguments by value

Question

The example code of what I am trying to ask is below.
None of the examples on the internet try to overload argument value as such.
One of the argument is a bool value and I want to overload a method based on the bool value rather than the usual, argument type.

from typing import overload, Union

@overload
def myfunc(arg:bool=True)-> str: ...

@overload
def myfunc(arg:bool=False)-> int: ...

def myfunc(arg:bool)->Union[int, str]:
    if arg: return "something"
    else: return 0

Is the overloading code in the above example code correct ?
Can you give an example/blog/source that mentions this kind of overloading, since I couldn't find anything in Python docs and pep-484

I found one probable way of doing it is with typing.Literal as used in latest python docs (since python v3.8)

from typing import overload, Union, Literal

@overload
def myfunc(arg:Literal[True])-> str: ...

@overload
def myfunc(arg:Literal[False])-> int: ...

def myfunc(arg:bool)->Union[int, str]:
    if arg: return "something"
    else: return 0

But I cannot move to python 3.8 just yet as I am working on production code that is still on python 3.6, soon upgrading to 3.7 at best.
Hence I am still looking for answers on how to achieve this in python 3.6

Answer 1

Building off the correct accepted answer, @overloading for default (keyword) arguments can be achieved as follows:

from typing import overload, Union, Literal

@overload
def myfunc(arg: Literal[True] = True) -> str: ...

@overload
def myfunc(arg: Literal[False]) -> int: ...

def myfunc(arg: bool =  True) -> Union[int, str]:
    if arg: return "something"
    else: return 0

The first overload (with Literal[True]) is used for type-checking function calls myfunc(True) and myfunc(), while the second overload (with Literal[False]) is used for type-checking the function call myfunc(False).