When is "i += x" different from "i = i + x" in Python?

Question

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?

Asked By: MarJamRob

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Answer 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.}

Answered By: mgilson

Answer 2

Under the covers, i += 1 does something like this:

try:
    i = i.__iadd__(1)
except AttributeError:
    i = i.__add__(1)

While i = i + 1 does something like this:

i = i.__add__(1)

This is a slight oversimplification, but you get the idea: Python gives types a way to handle += specially, by creating an __iadd__ method as well as an __add__.

The intention is that mutable types, like list, will mutate themselves in __iadd__ (and then return self, unless you’re doing something very tricky), while immutable types, like int, will just not implement it.

For example:

>>> l1 = []
>>> l2 = l1
>>> l1 += [3]
>>> l2
[3]

Because l2 is the same object as l1, and you mutated l1, you also mutated l2.

But:

>>> l1 = []
>>> l2 = l1
>>> l1 = l1 + [3]
>>> l2
[]

Here, you didn’t mutate l1; instead, you created a new list, l1 + [3], and rebound the name l1 to point at it, leaving l2 pointing at the original list.

(In the += version, you were also rebinding l1, it’s just that in that case you were rebinding it to the same list it was already bound to, so you can usually ignore that part.)

Answered By: abarnert

Answer 3

Here is an example that directly compares i += x with i = i + x:

def foo(x):
  x = x + [42]

def bar(x):
  x += [42]

c = [27]
foo(c); # c is not changed
bar(c); # c is changed to [27, 42]

Answered By: Deqing

When is "i += x" different from "i = i + x" in Python?

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