内置和用户定义类型之间的Python比较

Question

How does Python 3 compare a built-in object (on the lhs) to a user-defined object (on the rhs)?

Does the built-in __eq__ method simply delegate the comparison to the rhs ( rhs.__eq__(self) )?

I didn't find any statement about this in the docs. The docs state:

Objects of different types, except different numeric types, never compare equal.

It's quite misleading because:

class X:
  def __eq__(self, rhs)
    return True

x = X()
'abc' == x # True

I think the doc statement should be rephrased as follows:

Objects of different built-in types, except different numeric types, never compare equal.

and should furthermore clarify how the comparison to user-defined class instances is performed.

Answer 1

To answer the questions:

How does Python 3 compare a built-in object (on the lhs) to a user-defined object (on the rhs)?

The same way as with any other object comparisons (including None !).

Does the built-in __eq__ method simply delegate the comparison to the rhs ( rhs.__eq__(self) )?

No. The built-in __eq__ does not delegate like this. There is a higher-construct at work that covers the behavior of == in Python.

Given a == b , where a.__eq__(b) returns NotImplemented then b.__eq__(a) will be invoked and the result of the used as the result of the equality test. ( False is returned if both __eq__ implementations return NotImplemented .)

Thus, given x (of class X ), and given that "abc".__eq__(x) returns NotImplemented , then x.__eq__("abc") is invoked (and evaluates to True per the question).

The same applies to the other standard comparison operators.

While I don't care to speculate too much on the documentation (or possible mis-wording), I believe it is entirely accurate if taken in context of stdObj.__eq__(obj) as opposed to stdObj == obj .

See also:

• The documentation on the __eq__ protocol which briefly mentions NotImplemented
• Why return NotImplemented instead of raising NotImplementedError