Python inner function sets vs nonlocal

Question

I believe I know the answer to this, but wanted to double-check because I find this a bit confusing.

def outerFunc():
    mySet = set()
    a = 0
    def innerFunc():
        mySet.add(1)
        mySet.add(2)
        a = 7
    innerFunc()
    print(mySet) # {1, 2}
    print(a) # 0

Here, if I want to change the value of a, I need to use nonlocal. The fact that the set changes is just because sets are passed by reference? So, in an inner function we have access to the values of the outer function’s variables, but cannot modify them unless they’re references?

Asked By: adinutzyc21

||

Source

Answer 1

You can check the python document

In Python, variables that are only referenced inside a function are
implicitly global. If a variable is assigned a value anywhere within
the function’s body, it’s assumed to be a local unless explicitly
declared as global.

So if you assigned a variable and the variable without global just affects the local.
For example, if you assigned value to mySet, then it also does not change.

def outerFunc():
    mySet = set()   
    def innerFunc():
        mySet = {1}
        mySet.add(2)
    innerFunc()
    print(mySet)  # ''

Answered By: Ben

Python inner function sets vs nonlocal

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