def func(data): data = [7, 23, 42] print('Function scope: ', data) data = ['Peter', 'Paul', 'Mary'] print('Outer scope: ', data)

Question

Options

• AFunction scope: ['Peter', 'Paul', 'Mary'] Outer scope: ['Peter', 'Paul', 'Mary']
• BFunction scope: [7, 23, 42] Outer scope: ['Peter', 'Paul', 'Mary']
• CNone of the above.
• D1 Function scope: [7, 23, 42] 2 Outer scope: [7, 23, 42]

• A
  16% (6)
• B
  76% (28)
• C
  5% (2)
• D
  3% (1)

Explanation

Option B is correct because inside func, the line data = [7, 23, 42] creates a new local variable named data - it does not modify the outer data. Python's scoping rules (LEGB: Local, Enclosing, Global, Built-in) mean that any assignment inside a function creates a local binding, leaving the outer data = ['Peter', 'Paul', 'Mary'] completely untouched.

Why the distractors are wrong:

• A is wrong because it assumes the function somehow printed the outer list - but the local assignment data = [7, 23, 42] overrides the parameter inside the function's scope.
• D is wrong because the outer scope never gets reassigned; the function's local data dies when the function returns, so the outer print still sees ['Peter', 'Paul', 'Mary'], not [7, 23, 42].
• C ("None of the above") is wrong because B is a valid and correct output.

Memory tip: Think of a function as a "Vegas rule" - what happens inside the function, stays inside the function. Assigning a variable inside a function never reaches outside unless you explicitly use global or return the value.

No community discussion yet for this question.