How is __eq__ handled in Python and in what order?

Since Python does not provide left/right versions of its comparison operators, how does it decide which function to call?

class A(object):
    def __eq__(self, other):
        print "A __eq__ called"
        return self.value == other
class B(object):
    def __eq__(self, other):
        print "B __eq__ called"
        return self.value == other

>>> a = A()
>>> a.value = 3
>>> b = B()
>>> b.value = 4
>>> a == b
"A __eq__ called"
"B __eq__ called"
False

This seems to call both __eq__ functions.

I am looking for the official decision tree.

Solution 1:

The a == b expression invokes A.__eq__, since it exists. Its code includes self.value == other. Since int's don't know how to compare themselves to B's, Python tries invoking B.__eq__ to see if it knows how to compare itself to an int.

If you amend your code to show what values are being compared:

class A(object):
    def __eq__(self, other):
        print("A __eq__ called: %r == %r ?" % (self, other))
        return self.value == other
class B(object):
    def __eq__(self, other):
        print("B __eq__ called: %r == %r ?" % (self, other))
        return self.value == other

a = A()
a.value = 3
b = B()
b.value = 4
a == b

it will print:

A __eq__ called: <__main__.A object at 0x013BA070> == <__main__.B object at 0x013BA090> ?
B __eq__ called: <__main__.B object at 0x013BA090> == 3 ?

Solution 2:

When Python2.x sees a == b, it tries the following.

• If type(b) is a new-style class, and type(b) is a subclass of type(a), and type(b) has overridden __eq__, then the result is b.__eq__(a).
• If type(a) has overridden __eq__ (that is, type(a).__eq__ isn't object.__eq__), then the result is a.__eq__(b).
• If type(b) has overridden __eq__, then the result is b.__eq__(a).
• If none of the above are the case, Python repeats the process looking for __cmp__. If it exists, the objects are equal iff it returns zero.
• As a final fallback, Python calls object.__eq__(a, b), which is True iff a and b are the same object.

If any of the special methods return NotImplemented, Python acts as though the method didn't exist.

Note that last step carefully: if neither a nor b overloads ==, then a == b is the same as a is b.

From https://eev.ee/blog/2012/03/24/python-faq-equality/