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

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

CodePudding user response：

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

CodePudding user response：

In Both examples: foo2() is a function which returns a value.(i.e. 10) So, In first you catched that value and stored it in rv and at the same time res is incremented by 10 so now value of res=10 and rv=10 therefore In 1st example output is 20

Whereas in 2nd example, foo2() is called after the res is 0 so res=0 10 therefore In 2nd example output is 10