What is the expected output of the following code? def fun(n): n **= n return n print(fun(3))

Question

Options

• A9
• BThe program will cause an error.
• CTrue
• D27
• E3

• A
  12% (8)
• B
  3% (2)
• C
  8% (5)
• D
  76% (50)
• E
  2% (1)

Explanation

Option D is correct because **= is Python's exponentiation assignment operator, making n **= n equivalent to n = n ** n - so with n = 3, this computes 3 ** 3 = 27, which is then returned and printed.

Why the distractors fail:

• A (9) is the result of 3 * 3 or 3 ** 2 - a common confusion mistaking ** (exponentiation) for * (multiplication), or misreading the exponent as 2 instead of 3.
• B (error) is wrong because the code is syntactically and logically valid - Python handles **= without issue.
• C (True) would only appear if a boolean comparison (e.g., ==) were involved; no such comparison exists here.
• E (3) would be correct only if the function returned n before modifying it, but the modification happens first.

Memory tip: Read **= as "raise to the power of itself" - n **= n means n becomes n-to-the-n, so always ask: "base and exponent are the same number." For 3, that's 3³ = 27, not 3².

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