New Enroll in our GIC Finance Online Course Here!

    Register

    Question

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

    snippet, considering scope rules? x = 5                  def func():     x = 10     def inner():         nonlocal x         x += 1         return x     return inner()   print(func())
    A 6 Correct Answer Incorrect Answer
    B 11 Correct Answer Incorrect Answer
    C 10 Correct Answer Incorrect Answer
    D UnboundLocalError Correct Answer Incorrect Answer
    E None Correct Answer Incorrect Answer

    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.
    The nonlocal keyword ensures that changes to x in the inner() function persist in the func() scope. Without it, Python would treat x in inner() as a new local variable, causing an UnboundLocalError. Why Other Options Are Incorrect:
    • 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.

    Practice Next