Python Coding challenge - Day 483| What is the output of the following Python Code?

 

Code Explanation:

1. Global Scope (x = 10)

x = 10

The first x = 10 creates a variable x in the global scope (outside any function).

This value of x is available throughout the entire program unless overridden in a local scope.

2. Defining outer()

def outer():

    x = 20

    def inner(): nonlocal x; x = 30

    inner(); print(x)

Inside outer(), there's another variable x = 20, which is local to outer(). This means that x inside the outer() function initially takes the value 20.

Defining the inner() function:

def inner(): nonlocal x; x = 30

inner() is a nested function inside outer().

The nonlocal x statement tells Python to look for x in the nearest enclosing scope (which is outer()), rather than creating a new x in inner().

The x = 30 changes the value of x in the enclosing outer() function to 30.

Calling inner():

inner(); print(x)

When inner() is called, it updates the value of x in the outer() function to 30.

After inner() runs, x inside outer() is now 30.

The print(x) inside outer() will print the value of x from the outer() scope, which is now 30.

3. Calling outer()

outer(); print(x)

Executing outer():

outer() is called, and inside it, x becomes 20 initially.

inner() runs and modifies x to 30.

So, the first print statement inside outer() prints 30.

The final print(x):

This print statement is outside outer(), so it refers to the global x.

The global x was never modified by outer(). It still holds the value 10.

Hence, this print statement prints 10.

Final Output

30

10