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Question: What is the output of the following Python code snippet?

def append_to_list(value, item_list=[]):
    item_list.append(value)
    return item_list

list_a = append_to_list(10)
print(list_a)

list_b = append_to_list(20)
print(list_b)

Options:

    
• A) [10] followed by [20]
    
• B) [10] followed by [10, 20]
    
• C) [10] followed by [20, 10]
    
• D) TypeError: default argument cannot be mutated

Correct Answer: B) [10] followed by [10, 20]

Detailed Explanation:

This question tests your understanding of Python's handling of mutable default arguments.

    
• Function Definition Time: In Python, default parameter values are evaluated only once at the time the function is defined, not each time the function is called. Because a list ([]) is a mutable object, the same list instance persists across multiple calls.
    
• First Call (append_to_list(10)): Since no second argument is passed, Python uses the default list created at definition time. The number 10 is appended to it, and it returns [10].
    
• Second Call (append_to_list(20)): Again, no second argument is passed, so Python references the exact same list in memory that already contains 10. Appending 20 updates the list to [10, 20].

How to avoid this issue:

To ensure a new list is created on every call, use None as the default argument value:

def append_to_list(value, item_list=None):
    if item_list is None:
        item_list = []
    item_list.append(value)
    return item_list