Why do operators and methods of built-in types differ

[Csaba Hoch]

Hi,

if I write the following:
2

it seems to be exactly equivalent to this:
2

However, if I write invalid code and try to add a list to an int, the
errors will be different:

>>> 1+[]

Traceback (most recent call last):

>>> (1).__add__([])

NotImplemented

I found that operator.__add__(1, []) gives the same result as 1+[].

What is the reason behind this difference between the __add__ operator
and int.__add__?

Thank you,
Csaba

 

[andrew cooke]

What is the reason behind this difference between the __add__ operator
and int.__add__?

this is quite common in python. the special methods like __add__ are
used to implement some functionality (like '+' in this case), but they
are not all of it. for example, when
a + b
is evaluated, a.__add__(b) is attempted, but if that fails (raises a
NotImplemented error) then b.__radd__(a) is tried instead.

so there's not a 1-to-1 correspondence between '+' and __add__() and
that is reflected in the exceptions, too. when a method does not
exist a NotImplemented error is raised, but '+' contains extra logic
and raises a more useful error message.

does that make sense? i should probably add that this is just how i
understand things - i assume it's correct, but i've not looked
anything up in the documentation.

andrew