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GATE DA 2026 | Python & DS | Memory Based | Question 46

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1 1 vote

Consider the following Python code:

def append_to_lst(val, lst=[]):
    lst.append(val)
    return lst

print(append_to_lst(1))
print(append_to_lst(2))
print(append_to_lst(3, []))

What will be the output?

  1. $\verb|[1]|$
    $\verb|[2]|$
    $\verb|[3]|$
     
  2. $\verb|[1]|  $
    $\verb|[1, 2]|  $
    $\verb|[3]|$
     
  3. $\verb|[1]|  $
    $\verb|[2]|  $
    $\verb|[1, 2, 3]|$
     
  4. $\verb|[1]|  $
    $\verb|[1, 2]|  $
    $\verb|[1, 2, 3]|$

1 Answer

1 1 vote

Answer: B

 

The key to this problem is understanding how default arguments are handled in Python. When lst=[] is used as a default argument, the same list object is created only once when the function is defined, not every time the function is called. Subsequent calls to the function without providing a new list will reuse and modify this single, shared list object. 

  1. The first call, append_to_lst(1), uses the default list. The list becomes [1], and [1] is printed.
  2. The second call, append_to_lst(2), again uses the same default list object. The value 2 is appended to the existing list, making it [1, 2], and [1, 2] is printed.
  3. The third call, append_to_lst(3, []), explicitly provides a new, empty list ([]) as an argument. The value 3 is appended to this new list, making it [3], and [3] is printed.

The output will therefore be:

[1]
[1, 2]
[3]

 

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