In Python, what will be the output of the following code

Solution

The code demonstrates the use of the nonlocal keyword in Python, which allows a variable defined in an enclosing (but not global) scope to be modified. Here’s a breakdown:

1. The variable x = 5 is declared globally.
2. Inside the func() function, a local x = 10 is declared.
3. Within the nested inner() function, the nonlocal x statement is used. This tells Python that x refers to the x variable in the enclosing func() scope (not the global scope).
4. The x is incremented by 1 (x += 1), updating it to 11.
5. The inner() function returns x, and func() also returns the result from inner(). Hence, the final output is 11.

• Option A (6): This assumes that nonlocal affects the global x. However, nonlocal only modifies variables in the closest enclosing non-global scope. Thus, the global x = 5 remains unchanged.
• Option C (10): This would be true if no modification occurred to the local x inside the inner() function. However, x is incremented by 1 due to nonlocal.
• Option D (UnboundLocalError): This error would occur if nonlocal was omitted and x was incremented without declaration, as Python would not recognize it as being defined in the current scope.
• Option E (None): This would result if func() or inner() explicitly returned None. Since the function returns x, this is incorrect.