What does the operator += return in Python

Question

Original Question

While I was trying to answer another person's question on stackoverflow about the difference between = and += in Python, I encountered the following problem:

class Foo:
    def __init__(self, value, name="default"):
        self.value = value
        self.name = name
        
    def __add__(self, that):
        return Foo(self.value + that.value)
    
    def __iadd__(self, that):
        self.value = self.value + that.value
        return self
    
    def __str__(self):
        return "name: {}, value: {:d}".format(self.name, self.value)
    
a = Foo(1, 'alice')
b = Foo(2, 'bob')
print(a+=b)

The last print call was not successful and gave me this:

File "<ipython-input-8-0faa82ba9e4a>", line 3
    print(a+=b)
            ^
SyntaxError: invalid syntax

I don't know why this isn't working. Maybe it has something to do with the keyword argument passing mechanism? I just can't find any resource on this topic, since the overloaded __iadd__ method already returns a Foo object.

************** update ******************

If I change the __iadd__ method like this (just remove the return statement):

...
    def __iadd__(self, that):
        print("__iadd__ was called")
        self.value = self.value + that.value

a = Foo(1, 'alice')
b = Foo(2, 'bob')
a += b
print(a)  # Outputs: None

So, the final return statement in __iadd__ is indeed required. But it does not function as I thought (I come from a C/C++ background, so this behavior is a bit strange for me)

************************* 2nd Update ********************************

I almost forget that = in Python makes up a statement instead of an expression. The return statement in __iadd__ and my experience with other languages gives me an illusion that += could be used as an expression.

As stated in the Python documentation, __add__ is used to construct a new object. __iadd__ is designed for inplace modifications. But it all depends on the implementation. Although __iadd__ returns a object, this object is somehow "intercepted" by the language and reassigned to the left-hand operator, and, the final effect is, __iadd__ remains a statement, not an expression. Please correct me if I'm wrong. Also, I didn't find any resource confirming that += is a statement.

Summary

1. A plain code, say a = 1 is an assignment statement. It's not allowed to be used as a function argument. However, keyword argument passing is not restricted by that: print('Hello world', end='') still works.
2. • x = x + y is equivalent to x = x.__add__(y),
   • x += y is equivalent to x = x.__iadd__(y), check the doc for more details.
3. An example:

class Foo:
    def __init__(self, value, name="default"):
        self.value = value
        self.name = name

    def __add__(self, that):
        return Foo(self.value + that.value)

    def __iadd__(self, that):
        self.value = self.value + that.value
        return self

    def __str__(self):
        return "name: {}, value: {:d}".format(self.name, self.value)

a = Foo(1, 'alice')
b = Foo(2, 'bob')
c = a + b # objects a and b are unchanged
a += b    # object a is changed
print(c) # name: default, value: 3
print(a) # name: alice, value: 3

Answer 1

Assignments/in-place operations can't be considered as expressions that can be passed on, and that's by design.

Among other things, Guido wanted to avoid the infamous C typo

if (a=b)  // user actually wanted to test a==b

But python 3.8 provides a new assignment method known as assignment expression that allows to pass the assigned value as a parameter (which is very handy in list comprehensions).

You still cannot do in-place add but you can do add then assign:

>> a=2
>> b=3
>> print(a:=a+b)
5