TypeError after overriding the __add__ method

Question

I am trying to understand how __add__ works:

class MyNum:
    def __init__(self,num):
        self.num=num
    def __add__(self,other):
                 


        

Answer 1

You need to define __radd__ as well to get this to work.

__radd__ is reverse add. When Python tries to evaluate x + y it first attempts to call x.__add__(y). If this fails then it falls back to y.__radd__(x).

This allows you to override addition by only touching one class. Consider for example how Python would have to evaluate 0 + x. A call to 0.__add__(x) is attempted but int knows nothing about your class. You can't very well change the __add__ method in int, hence the need for __radd__. I suppose it is a form of dependency inversion.

As Steven pointed out, sum operates in place, but starts from 0. So the very first addition is the only one that would need to use __radd__. As a nice exercise you could check that this was the case!

Answer 2

>>> help(sum)
Help on built-in function sum in module __builtin__:

sum(...)
    sum(sequence[, start]) -> value

    Returns the sum of a sequence of numbers (NOT strings) plus the value
    of parameter 'start' (which defaults to 0).  When the sequence is
    empty, returns start.

In other words, provide a start value:

sum(d, MyNum(0))

Edit pasted from my below comment:

sum works with a default start value of the integer zero. Your MyNum class as written does not know how to add itself to integers. To solve this you have two options. Either you can provide a start value to sum that has the same type as you class, or you can implement __radd__, which Python calls when adding values of differing types (such as when the first value in d is added to the default start value of zero).

Answer 3

Another option is reduce (functools.reduce in Python 3.x).

from functools import reduce
from operators import add
d=[MyNum(i) for i in range(10)]
my_sum = reduce(add,d)

Answer 4

I oppose relaying on sum() with a start point, the loop hole exposed below,

In [51]: x = sum(d, MyNum(2))

In [52]: x.num
Out[52]: 47

Wondering why you got 47 while you are expecting like …start from 2nd of MyNum() while leaving first and add them till end, so the expected result = 44 (sum(range(2,10))

The truth here is that 2 is not kept as start object/position but instead treated as an addition to the result

sum(range(10)) + 2

oops, link broken !!!!!!

Use radd

Here below the correct code. Also note the below

Python calls __radd__ only when the object on the right side of the + is your class instance eg: 2 + obj1

#!/usr/bin/env python

class MyNum:
    def __init__(self,num):
        self.num=num

    def __add__(self,other):
        return MyNum(self.num+other.num)

    def __radd__(self,other):
        return MyNum(self.num+other)

    def __str__(self):
        return str(self.num)

d=[MyNum(i) for i in range(10)]
print sum(d)    ## Prints 45
d=[MyNum(i) for i in range(2, 10)]
print sum(d)    ## Prints 44
print sum(d,MyNum(2))   ## Prints 46 - adding 2 to the last value (44+2)

Answer 5

class MyNum:
    def __init__(self,num):
        self.num=num
    def __add__(self,other):
        return self.num += other.num
    def __str__(self):
        return str(self.num)

one = MyNum(1)
two = MyNum(2)

one + two

print(two.num)