How does the sum assignment operator work internally?

Question

I am creating a class and I want its objects to be summed up, using the sum operator or the sum assignment operator. I know that to use the sum operator, we must create for the class the special method __add__, as in the code below:

class list(object):
    def __add__(self, value, /):
        # Implementação de list

[1, 2, 3] + [4, 5, 6] # Utilizando o operador de soma, o retorno será: [1, 2, 3, 4, 5, 6]

But for the operator of sum assignment? Do you use this same method? I’m asking this question because I can’t understand how this operator works internally.

I even thought about the possibility of the method __add__ return the modified object itself, but it seems that this is not what happens. If we create, for example, a list and use that operator to add it with another one, the result will be the same list.

lista = [1, 2, 3]
id(lista) # 39196952
lista += [4, 5, 6]
id(lista) # 39196952  <- É a mesma lista. Houve uma modificação diretamente no objeto.

But if we use only the sum operator, the result is a new list (a new object):

lista = [1, 2, 3]
id(lista)      # 39196952
lista_nova = lista + [4, 5, 6]
id(lista_nova) # 39241608  <- What!? Ele retornou um novo objeto com a soma realizada.

Clearly there is a difference in the way in which the two operators work internally. That said, I would like to know:

• How the sum assignment operator works internally?
• The method __add__ has connection with that operator? If not, which method to use?

Answer 1

Operators + and += have different implementations when you use += (__iadd__) on an object of the type list the original list is mutated, so the memory reference remains. Already the operator + (__add__) generates a new object of type list and returns it.

You can read more about it in this question here.