How are Python in-place operator functions different than the standard operator functions?

Question

From the docs:

Many operations have an “in-place” version. The following functions provide a more primitive access to in-place operators than the usual syntax does; for example, the statement x += y is equivalent to x = operator.iadd(x, y). Another way to put it is to say that z = operator.iadd(x, y) is equivalent to the compound statement z = x; z += y.

Questions:

Why isn't operator.iadd(x, y) equivalent to z = x; z += y?
How does operator.iadd(x, y) differ from operator.add(x, y)?

Related question, but I'm not interested in Python class methods; just regular operators on built-in Python types.

Jegschemesch · Accepted Answer · 2012-02-23 11:40:18Z

32

First, you need to understand the difference between __add__ and __iadd__.

An object's __add__ method is regular addition: it takes two parameters, returns their sum, and doesn't modify either parameter.

An object's __iadd__ method also takes two parameters, but makes the change in-place, modifying the contents of the first parameter. Because this requires object mutation, immutable types (like the standard number types) shouldn't have an __iadd__ method.

a + b uses __add__. a += b uses __iadd__ if it exists; if it doesn't, it emulates it via __add__, as in tmp = a + b; a = tmp. operator.add and operator.iadd differ in the same way.

To the other question: operator.iadd(x, y) isn't equivalent to z = x; z += y, because if no __iadd__ exists __add__ will be used instead. You need to assign the value to ensure that the result is stored in both cases: x = operator.iadd(x, y).

You can see this yourself easily enough:

import operator
a = 1
operator.iadd(a, 2)
# a is still 1, because ints don't have __iadd__; iadd returned 3

b = ['a']
operator.iadd(b, ['b'])
# lists do have __iadd__, so b is now ['a', 'b']

edited Feb 23, 2012 at 11:40

Jegschemesch

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answered Jan 23, 2011 at 9:21

Glenn Maynard

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3 Comments

lig Over a year ago

docs.python.org/2/reference/datamodel.html#object.__iadd__ Notice that augmented assignments could but does not have to modify object in-place

max Over a year ago

Why is python designed with augmented assignments operating in-place? It's kinda feels counter-intuitive because when we learn the language we usually learn augmented assignments with immutable values first. So it's a bit unexpected that with mutable values augmented assignment behaves differently. Feels rather implicit and bug-prone.

juanpa.arrivillaga Over a year ago

@max that sounds like the fault of whoever designed the learning materials. The purpose of augmented assignment is quite clear: PEP 203

Leftium · Accepted Answer · 2011-01-23 09:54:58Z

0

Perhaps because some Python objects are immutable.

I'm guessing operator.iadd(x, y) is equivalent to z = x; z += y only for mutable types like dictionaries and lists, but not for immutable types like numbers and strings.

edited Jan 23, 2011 at 9:54

answered Jan 23, 2011 at 8:47

Leftium

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