🔍 GUESS THE OUTPUT #6 - full breakdown

python
def outer():
    x = 10
    def inner():
        print(x)
        x += 1
    inner()

outer()

Answer: UnboundLocalError: local variable 'x' referenced before assignment

This one surprises a lot of people - it does NOT print 10. Here's why: because inner() assigns to x (x += 1), Python treats x as a LOCAL variable throughout the entire function the moment it sees any assignment to it, even before that line executes. So the print(x) line is trying to read a local x that hasn't been assigned yet.

Fixed version:

python
def outer():
    x = 10
    def inner():
        nonlocal x
        print(x)
        x += 1
    inner()

The nonlocal keyword tells Python "this x refers to the enclosing function's variable, not a new local one."

This is a genuinely tricky one - Python decides variable scope at compile time based on whether a name is assigned anywhere in the function, not based on execution order. Great interview question for testing real understanding of closures vs. just pattern-matching syntax.