Delete __iadd__ and everything worked the same...

This can be a little confusing at first.

__iadd__ is the "inplace addition" operator. It is different than the __add__ method which simply returns the RESULT of the addition to another object. But instead the __iadd__ is 2 operations with one operator... it evaluates first then and replaces itself with a new object.

Since the addition is "inplace" it means that the value of object is modified in place. Its counterparts are __isub__ and iconcat for strings.

x = 3
# the iadd adds and modifies the object, 2 operations in one!
x += 1
print(x)

I agree with Matthew Ison. I also found the definition of iadd to not be necessary. Jeff, it is not that Matthew and I do not understand the purpose of iadd (i.e. addition and then assignment to the original variable), it is that in practice we have found it to not be necessary. Perhaps, we are using a later version of Python?

I haven't read Chris' response @chrisfreeman3, he's a smart guy for sure. Returning the correct type is a good use case for this.

Make sure you note this, as a programming interview may try to trip you up with this.

For practical purposes, __iadd__ makes the += operation work CORRECTLY-- that is, it DOES NOT need to create a NEW copy of the variable. The one that only implements the __add__ ends up replacing the original instance.

Think of it like this: normally, when you use the + operator (handled by the __add__ method), it’s as if you’ve mixed two things together in a bowl and poured them into a fresh, new container. But with __iadd__, Python is much more practical. Instead of fetching a new container, it just says, "Right, let’s keep this one. We’ll modify it right here." In terms of everyday life, it's like adding a few more steps to your pedometer without buying a whole new one each time. It’s all very efficient.

In many cases, the __add__ operator is just fine, but edge cases you would want the __iadd__.

Does this matter? Sometimes.

TL;DR proof

class Point1:
    # This is an implementation of the Point class, it can add objects to a third object
    # but it is missing the ability to ACTUALLY modify the object in-place
    def __init__(self, x, y):
        self.x = x
        self.y = y

    def __add__(self, other):
        # Adding the corresponding coordinates
        return Point1(self.x + other.x, self.y + other.y)

    def __repr__(self):
        return f"Point1({self.x}, {self.y})"

class Point2:
    # This is the original implementation of the Point class, it can modify the object in-place.
    def __init__(self, x, y):
        self.x = x
        self.y = y

    def __add__(self, other):
        # Adding the corresponding coordinates
        return Point2(self.x + other.x, self.y + other.y)

    def __iadd__(self, other):
        # Modifying the object in place by adding the coordinates
        self.x += other.x
        self.y += other.y
        return self  # Returning the modified object itself

    def __repr__(self):
        return f"Point2({self.x}, {self.y})"



# Example usage
p1 = Point1(2, 3)
print(f'Original: p1 = {p1}, id(p1): {id(p1)}')
p2 = Point1(4, 1)
p1 += p2  # This calls p1.__add__(p2)

# Note the id, it is different than the original
print(f'Inplace operation: {p1}, id(p1): {id(p1)}')


# Example usage
q1 = Point2(2, 3)
print(f'Original: q1 = {q1}, id(q1): {id(q1)}')
q2 = Point2(4, 1)
q1 += q2  # This calls p1.__iadd__(q2)

# Note the id, it is using the SAME memory location.
print(f'Inplace operation: {q1}, id(q1): {id(q1)}')

Here is a run of the program. It clearly shows the __add__ method produces a NEW memory location when used. The second with the __iadd__ method shows it used the SAME location.

(base) jeff@pop-os:~/projects/treehouse-dev$ python3 test_iadd.py 
Original: p1 = Point1(2, 3), id(p1): 129052629523040
Inplace operation: Point1(6, 4), id(p1): 129052629523136
Original: q1 = Point2(2, 3), id(q1): 129052629523040
Inplace operation: Point2(6, 4), id(q1): 129052629523040