consider the following code:
>>> x = y = [1, 2, 3, 4]
>>> x += [4]
>>> x
[1, 2, 3, 4, 4]
>>> y
[1, 2, 3, 4, 4]
and then consider this:
>>> x = y = [1, 2, 3, 4]
>>> x = x + [4]
>>> x
[1, 2, 3, 4, 4]
>>> y
[1, 2, 3, 4]
Why is there a difference these two?
(And yes, I tried searching for this).
__iadd__ mutates the list, whereas __add__ returns a new list, as demonstrated.
An expression of x += y first tries to call __iadd__ and, failing that, calls __add__ followed an assignment (see Sven's comment for a minor correction). Since list has __iadd__ then it does this little bit of mutation magic.
__iadd__() or __add__() is called. list.__iadd__() simply returns self, though, so the assignment has no effect other than rendering the target name local to the current scope.
__iadd__ works faster for lists (it changes object in-place instead of recreating one); it allows to append any iterable, a = [] a+=range(5) works for lists, and a = a + smth requires smth to be an instance of list
__add__ and __iadd__ behave so surprisingly differently.The first mutates the list, and the second rebinds the name.
1)'+=' calls in-place add i.e iadd method. This method takes two parameters, but makes the change in-place, modifying the contents of the first parameter (i.e x is modified). Since both x and y point to same Pyobject they both are same.
2)Whereas x = x + [4] calls the add mehtod(x.add([4])) and instead of changing or adding values in-place it creates a new list to which a points to now and y still pointing to the old_list.
