__iadd__

Good question! You are correct, in Kenneth's code, with or without __iadd__ you get the same result: an int. This is arguably not the correct result. Ideally, an __iadd__ method should keep the type of object the same. As mentioned in the Python docs:

These [augmented arithmetic assignments] methods should attempt to do the operation in-place (modifying self) and return the result (which could be, but does not have to be, self).

So the __iadd__ method should look more like:

    def __iadd__(self, other):
        print(f"{self.value} running __iadd__ with value: {other}")
        self.value = str(self + other)
        return self

Since Kenneth's version of __iadd__ sets self.value to self + other, it is now an int type instead of a str type and the returned int value is assigned to the original value. This is the same result as if __add__ was used instead.

To complete the options, if __iadd__ is not implemented, then __add__ is used as a work-around:

a += b
# becomes
tmp = a + b
a = tmp

Post back if you have more questions. Good luck!!