Suppressing add when radd is defined

Question

I have two classes

Foo

and

Bar

:

class Foo(str):

    def __add__(self, other):

        return 'Foo.__add__ called'





class Bar(str):

    def __radd__(self, other):

        return 'Bar.__radd__ called'

As you can see, I have

Foo

implementing

__add__

and

Bar

implementing

__radd__

.

I can create objects of these classes and add them, because the magic methods are defined.

f = Foo('foo')

b = Bar('bar')

However, adding them produces this:

In [390]: f + b

Out[390]: 'Foo.__add__ called'

How can I set it so that

Bar.__radd__

method is called in preference to

Foo.__add__

?

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Answer 1

There are two ways you can do this.

Add an explicit check in Foo related to Bar.

class Foo(str):
    def __add__(self, other):
        if isinstance(other, Bar):
            return NotImplemented
        return 'Foo.__add__ called'


class Bar(str):
    def __radd__(self, other):
        return 'Bar.__radd__ called'

Make Bar a subclass of Foo.

From docs:

Note: If the right operand’s type is a subclass of the left operand’s type and that subclass provides the reflected method for the operation, this method will be called before the left operand’s non-reflected method. This behavior allows subclasses to override their ancestors’ operations.

Suppressing __add__ when __radd__ is defined

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Suppressing add when radd is defined