variable is redefined in augment assignment if it is used in closure

Question

Here is the example:

def foo1():
    res = 0
    def foo2():
        nonlocal res
        res += 10
        return 20

    rv = foo2()
    res += rv

    print(res)

foo1()  # output: 30 

res is increased by 10 in closure, becoming 10, then plus 10 returned by foo2, resulting in 20 consequently. So far, so good.

Here is an another similar case:

def foo1():
    res = 0
    def foo2():
        nonlocal res
        res += 10
        return 20

    res += foo2()

    print(res)

foo1()  # output: 20

Strangely, it outputs 10.

It seems that res is re-defined in res += foo2(). Do I think right?

Why is the result 10 at above example? what results in this?

• CPython 3.6.9

Answer 1

Based on the symptom, in stack pseudo-code it is acting like

push res, push foo2(), add, pop res

So the change to res in the middle of foo2 has no effect on the result. Because it "grabbed" the value of res BEFORE foo2, and used that in the expression.

Or you could show the pseudo-code like this:

temp1 = res
temp2 = foo2()
temp1 = temp1 + temp2
res = temp1