Operator binding in Python

Question

I have a containery object something like this:

class foo(object):
  def __init__(self,value):
    self.value = value
  def __eq__(self,other):
    return self.value == other

Which I can use like this:

>>> var = foo(4)
>>> var == 4
True
>>> var == 3
False
>>> # So far so good
>>> var == foo(4)
True
>>> var == foo(3)
False
>>> # Huh

My question is: what is happening that allows this to magically "just work"? How does Python bind the == operator inside __eq__ so that the value member of both sides is compared?

Not so strange? 3 != 4? var is effectively 4, so comparing it to anything other than 4 or foo(4) should return False? — Torxed
– Torxed, Commented Mar 26, 2014 at 14:39
it's operator overloading in python: ics.uci.edu/~pattis/ICS-33/lectures/operatoroverloading1.txt — venpa
– venpa, Commented Mar 26, 2014 at 14:39
There's whole official documentation on magic methods: docs.python.org/2/reference/datamodel.html#object.__eq__ — J0HN
– J0HN, Commented Mar 26, 2014 at 14:39

Martijn Pieters · Accepted Answer · 2014-03-26 14:47:28Z

6

If an __eq__ method returns NotImplemented for a given argument, Python will try the same test in reverse. So x == y will try x.__eq__(y) first, if that returns the NotImplemented singleton, y.__eq__(x) is tried.

This applies to the comparison inside your __eq__ method as well; you are essentially comparing:

self.value == other

which is:

4 == other

which is:

4.__eq__(other)

which returns NotImplemented:

>>> var = foo(4)
>>> (4).__eq__(var)
NotImplemented

So Python tries other.__eq__(4) instead, which returns True.

edited Mar 26, 2014 at 14:47

answered Mar 26, 2014 at 14:41

Martijn Pieters

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14 Comments

rophl Over a year ago

My local implementation of Python gives AttributeError: 'int' has no attribute '__eq__'. I guess any exception prompts the reversal of the comparison? Do similar things happen if a class implements __gt__ but not __lt__?

M4rtini Over a year ago

So when comparing var == foo(4) this get's reduced to var.value == foo(4) Which should raise a notimplemented error. Does it then try the reverse as foo(4) == var.value or foo(4) == var? would it try the second if the first also raised a notimplemented error?

mgilson Over a year ago

Note that things are slightly different for methods which have reflected bindings in the object model -- e.g. __add__. In that case, I believe the call order is __add__(x, y) -> __radd__(x, y)

Martijn Pieters Over a year ago

@M4rtini: var.__eq__(foo(4)) is called, this then executes another comparison, var.other == foo(4). Python tries that separately, so first var.other.__eq__(foo(4)) returning NotImplemented, then foo(4).__eq__(var.other), returning True. So the outermost var.__eq__(foo(4)) call does not return NotImplemented, it returns True.

Martijn Pieters Over a year ago

@M4rtini: Yes, if the foo.__eq__() method itself explicitly returns NotImplemented then other.__eq__() will be tried.

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