``python def test(x=1, y=2): x = x + y y += 1 print(x, y) test() `` What is the expected output of the following code?

Question

def test(x=1, y=2):
 x = x + y
 y += 1
 print(x, y)

test()

Options

• A13
• B33
• CThe code is erroneous.
• D11
• E31

• B
  78% (18)
• C
  4% (1)
• D
  13% (3)
• E
  4% (1)

Explanation

Option B (3 3) is correct because when test() is called with no arguments, the defaults x=1 and y=2 are used. The first line inside the function sets x = 1 + 2 = 3, then y += 1 increments y from 2 to 3, so print(x, y) outputs 3 3.

Why the distractors are wrong:

• A (1 3) - Assumes x was never updated, but x = x + y reassigns it to 3, not leaving it as 1.
• C (erroneous) - The code is perfectly valid Python; default parameters and in-place arithmetic are both legal.
• D (1 1) - Misreads both operations entirely; neither value stays at 1 after execution.
• E (3 1) - Gets x right but confuses y += 1 as subtracting or resetting; y goes from 2 to 3, not down to 1.

Memory tip: Trace assignments top-to-bottom and remember that default parameters are only the starting values - any reassignment inside the function immediately replaces them. Write out each variable's value after every line to avoid mix-ups.

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