How to use strongtyping-pyoverload on a class based level

Overloading methods

from strongtyping_pyoverload import overload


class Foo:

    @overload
    def func(self, a: str):
        print("Called with `str`")

    @overload
    def func(self, a: list):
        print("Called with `list`")

    @overload
    def func(self, a: tuple):
        print("Called with `tuple`")

>>> foo = Foo()
>>> foo.func("hello")
Called with `str`
>>> foo.func(list("hello"))
Called with `list`
>>> foo.func(tuple("hello"))
Called with `tuple`

Subclasses/Inheritance

can overwrite an existing method but these must match the exact type definition of the original method

from strongtyping_pyoverload import overload


class Example:
    @overload
    def other_func(self):
        return 0

    @overload
    def other_func(self, a: int, b: int):
        return (a * a) / b


class Other(Example):

    @overload
    def other_func(self, a):
        return a ** a + a

    @overload
    def other_func(self, a: int, b: int):  # the parameters and everything are exact the same
        return ((a * a) / b) + a

>>> example = Example()
>>> example.other_func(2, 3)
1.333333333333333
>>>
>>> other = Other()
>>> other.other_func()
0
>>> other.other_func(2)
6
>>> other.other_func(2, 3)
3.333333333333333

A type hint for each parameter??

you only need to have one typed parameter which differ or have a different length for your parameters

from strongtyping_pyoverload import overload


class Other:

    @overload
    def other_func(self, a: int, b):
        return (a + b) * (a + b)

    @overload
    def other_func(self, a: str, b):
        return f'{a.lower()}_{b.lower()}'

    @overload
    def other_func(self, a: list, b):
        return len(a) * b

    @overload
    def other_func(self, a, b, c):
        return a + b + c

>>> other = Other()
>>> other.other_func("Hello", "World")
hello_world
>>> other.other_func(2, 2)
16
>> > other.other_func([1, 2, 3], 2)
6
>>> other.other_func(2, 3, 4)
9

Variable Parameter length

you can also define the same function with different no. of parameters with the same type

from strongtyping_pyoverload import overload


class Other:

    @overload
    def other_func(self, a: int, b: int):
        return (a + b) * (a + b)

    @overload
    def other_func(self, a: int, b: int, c: int):
        return (a + b) * (a + b)

>>> other = Other()
>>> other.other_func(3, 4)
49
>>> other.other_func(3, 4, 5)
245

No function matches

when no function matches an AttributError will be raised

from strongtyping_pyoverload import overload


class Example:
    @overload
    def other_func(self, a: int, b: int):
        return (a + b) * (a + b)

>>> example = Example()
>>> example.other_func("Not", "Supported")
Traceback (most recent call last):
...
AttributeError: `Example` has no function which matches with your parameters `('Not', 'Supported')`