["python","operators"]
When is "i += x" different from "i = i + x" in Python?

I was told that += can have different effects than the standard notation of i = i +. Is there a case in which i += 1 would be different from i = i + 1?
This depends entirely on the object i. += calls the __iadd__ method (if it exists -- falling back on __add__ if it doesn't exist) whereas + calls the __add__ method1 or the __radd__ method in a few cases2. From an API perspective, __iadd__ is supposed to be used for modifying mutable objects in place (returning the object which was mutated) whereas __add__ should return a new instance of something. For immutable objects, both methods return a new instance, but __iadd__ will put the new instance in the current namespace with the same name that the old instance had. This is why i = 1 i += 1 seems to increment i. In reality, you get a new integer and assign it "on top of" i -- losing one reference to the old integer. In this case, i += 1 is exactly the same as i = i + 1. But, with most mutable objects, it's a different story: As a concrete example: a = [1, 2, 3] b = a b += [1, 2, 3] print(a) # [1, 2, 3, 1, 2, 3] print(b) # [1, 2, 3, 1, 2, 3] compared to: a = [1, 2, 3] b = a b = b + [1, 2, 3] print(a) # [1, 2, 3] print(b) # [1, 2, 3, 1, 2, 3] notice how in the first example, since b and a reference the same object, when I use += on b, it actually changes b (and a sees that change too -- After all, it's referencing the same list). In the second case however, when I do b = b + [1, 2, 3], this takes the list that b is referencing and concatenates it with a new list [1, 2, 3]. It then stores the concatenated list in the current namespace as b -- With no regard for what b was the line before. 1In the expression x + y, if x.__add__ isn't implemented or if x.__add__(y) returns NotImplemented and x and y have different types, then x + y tries to call y.__radd__(x). So, in the case where you have foo_instance += bar_instance if Foo doesn't implement __add__ or __iadd__ then the result here is the same as foo_instance = bar_instance.__radd__(bar_instance, foo_instance) 2In the expression foo_instance + bar_instance, bar_instance.__radd__ will be tried before foo_instance.__add__ if the type of bar_instance is a subclass of the type of foo_instance (e.g. issubclass(Bar, Foo)). The rationale for this is that Bar is in some sense a "higher-level" object than Foo so Bar should get the option of overriding Foo's behavior.