Convert custom objects to int with explicit base

Question

Let's say we have the following class:

class Foo:
    def __init__(self):
        self.bar = "12"

    def __str__(self):
        return self.bar
    
    def __int__(self, *args, **kwargs):
        print("inside __int__")
        return int(self.bar, *args, **kwargs)


a = Foo()
print(int(a))
# -> inside __int__
# -> 12

The result of printing this is as you would expect. However if we try:

print(int(a, 10))   # or any other valid base instead of 10

the following exception is raised:

TypeError: int() can't convert non-string with explicit base

Also, inside __int__ is not printed meaning that a.__int__ is not called. Is there any special method which would allow specifying a base in the int() call or do I just have to live with this behavior?

Even more interesting (and more in the direction of what I want to do) is the following:

class Foo(str):   # subclass of str and excepts argument in constructor
    def __init__(self, bar):
        self.bar = bar

    def __int__(self, *args, **kwargs):
        return int(self.bar, *args, **kwargs)


a = Foo("12")
print(int(a), int(a, 10))
# -> 12 12
# Actually works

a.bar = "34"
print(int(a), int(a, 10))
# -> 34 12
# what?

My assumption on this one is that int(num, base) uses num.__getnewargs__(). Either way, is there any way to get around these examples?

jonrsharpe · Accepted Answer · 2022-08-06 12:35:41Z

1

This behaviour is specifically called out in the documentation of int:

If x is not a number or if base is given, then x must be a string, bytes, or bytearray instance representing an integer literal in radix base.

The __int__ "magic method" (or __index__, __trunc__ which are used as fallbacks) is only invoked when the base argument is not provided.

answered Aug 6, 2022 at 12:35

jonrsharpe

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3 Comments

Mr_HxID Over a year ago

Yes, I've read the documentation on this. So in other words, it's up to me to say int(a.bar, some_base) instead, right?

jonrsharpe Over a year ago

@Mr_HxID that could be one way to handle it, yes. Or e.g. if the base was a property of the instance the __int__ implementation could return(self.bar, self.base) and be used with the single-arg int.

Mr_HxID Over a year ago

Ok, I'll probably do something similar to this then. Still, I find it kind of silly of python to place such a restriction on the multi-arg int. But thanks for your quick reply.

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