Here is a minimal reproducible example:

class Attribut:
    def __init__(
        self,
        name: str,
        other_name: str,
    ):
        self.name: str = name
        self.other_name: str = other_name
        
    def __eq__(self, other):
        if isinstance(other, Attribut):
            return self.name == other.name and self.other_name == other.other_name
        else:
            return NotImplemented
    
    def __hash__(self):
        return 0

If I try to do:

a = Attribut("lol", "a")
print(a==4)

I thought I would get NotImplemented, but instead I get False.

Also, what is the point of overriding __hash__, I cannot find a situation where this is useful?

CodePudding user response：

You have to raise an exception, not return it

def __eq__(self, other):
    if isinstance(other, Attribut):
        return self.name == other.name and self.other_name == other.other_name
    else:
        raise NotImplemented

Overriding hash method allows you to use other methods that would involve sorting a list of your Attribut objects.

To test it create a script in which you would create such a list

a = [Attribut("lol", "a"),  Attribut("lol", "a"),  Attribut("lol", "a")]

and then try to make a set out of it

set(a)

If the method hash is implemented in Attribut class then there would be no problem. If not then the program will return an error TypeError: unhashable type: 'Attribut'

CodePudding user response：

When a.__eq__(4) returns NotImplemented, you don't get the value back immediately. Instead, Python attempts to call the reflected version of __eq__ (__eq__ itself) with the other argument, namely (4).__eq__(a). It's this call that returns False.

By returning NotImplemented, you are not saying that self and other cannot be compared for equality (for that, raise a ValueError), but rather that Attribut.__eq__ does not know how to do so, but perhaps the other argument's __eq__ method does.