``python def func(x, y): if x == y: return x else: return func(x, y-1) print(func(0, 3)) `` What is the output of the following snippet?

Question

def func(x, y):
 if x == y:
 return x
 else:
 return func(x, y-1)

print(func(0, 3))

Options

• AThe snippet will cause a runtime error.
• B0
• C1
• D2

• A
  16% (7)
• B
  73% (33)
• C
  4% (2)
• D
  7% (3)

Explanation

Option B is correct because the function recursively decrements y by 1 on each call until x == y. Starting with func(0, 3), the chain goes func(0,3) → func(0,2) → func(0,1) → func(0,0), at which point x == y and the function returns x, which is 0.

Why the distractors are wrong:

• A is wrong because recursion terminates cleanly - y decrements toward x=0 and the base case is always reached.
• C and D are wrong because the return value is x (which never changes), not the intermediate value of y at any step.

Memory tip: Read the function as "keep subtracting 1 from y until y equals x, then return x." Since x is fixed at 0 and y chases it downward, the answer is always whatever x was passed in - here, 0.

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