What is being called when doing int() and float()?

Question

I understand that when I do:

>>> int(4.0)
>>> int('10')

The __init__() method from the class of the argument (in the example, float and str, respectively), will be called.

I wonder what will happen if I do int(), I mean, without any argument. It wont call __int__() from class NoneType, because it doesn't implement that method.

What will happen?

Also, does this mean I can only directly call the constructor of class int if the argument is an int?

Thanks,

Why do you think `int()` calls `__init__` on anything? (and technically speaking, `str` and `int` don't even have `__init__` methods). — Martijn Pieters, Aug 25 '16 at 14:54
Image the argument to `int` has a default value of 0: `def __new__(cls, value=0)`. — chepner, Aug 25 '16 at 14:56
Especially why you think it would call it on the *argument*, rather than on the `int` class. — Daniel Roseman, Aug 25 '16 at 14:57

score 1 · Accepted Answer · edited May 23 '17 at 12:34

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in python everything is an object, so when you call:

x = int(y) then constructor int.__new__(cls, value=y) is called.

this returns an int object and assigns it to x.

x may have a method bound to it: x.__init__() to control it's setup after creation, but creating the object in the first place is handled by the method __new__() of the base class.

If you provide no value to int(), the default value from the constructor call:

int.__new__(int, value=0) is substituted. Then as you mentioned, int has no __init__ method to modify it's setup after creation..

edited May 23 '17 at 12:34

Community

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answered Aug 25 '16 at 15:18

Aaron

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Thanks, that was it. Makes sense. – Tony Power Aug 25 '16 at 22:05

What is being called when doing int() and float()?

1 Answers1