3. Given the following function

def nPrint(message, n):
while n 0:
print(message)
n -= 1

What is k after invoking nPrint("A message", k)?

k = 2
nPrint(n = k, message = "A message")
A. 0
B. 1
C. 2
D. 3

Question

3. Given the following function

def nPrint(message, n):
while n > 0:
print(message)
n -= 1

What is k after invoking nPrint("A message", k)?

k = 2
nPrint(n = k, message = "A message")
A. 0
B. 1
C. 2
D. 3

BlackTom AI · Accepted Answer

The question presents a function nPrint(message, n) that loops while n > 0, printing the message and then decrementing n by 1.

We are then asked what value k has after calling nPrint(n = k, message = "A message"). The key details are:
- The function uses a parameter n as the loop counter, which is a local variable within the function scope.
- There is no return statement in nPrint, so it does not output a value or modify variables outside its scope.
- The call uses keyword arguments, assigning n = k and message = "A message". Given k is initially 2, this means the function starts with a local n value of 2 and prints twice before exiting.

Now, evaluating each option:

Option A (0): This would only occur if k were somehow set to 0 by the function, or if the function returned a value that replaced k. Since there is no return and the function’s local n is decremented from 2 to 1 to 0, k itself is unaffected. Therefore 0 is incorrect.

Option B (1): If the function modified k directly or returned a value that changed k, it could be 1. But the parameter n is a local copy; the loop runs twice (for n = 2 and n = 1) and ends, with no effect on k. Thus 1 is incorrect.

Option C (2): This matches the initial value of k and remains unchanged after the call because nPrint does not return anything or assign to k. The outer variable k stays 2, so this is the correct interpretation.

Option D (3): There is no mechanism in the function to produce or propagate a 3. The loop runs exactly two times given the starting n = 2, so 3 is incorrect.

In summary, the function operates on a local copy of n and has no side effects on the outer variable k. Therefore, after the call, k remains 2, which corresponds to option C.

3. Given the following function def nPrint(message, n): while n > 0: print(message) n -= 1 What is k after invoking nPrint("A message", k)? k = 2 nPrint(n = k, message = "A message") A. 0 B. 1 C. 2 D. 3单项选择题

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