Tuesday 19 March 2024

Python Coding challenge - Day 152 | What is the output of the following Python Code?

 

def foo(x, y=[]):

    y.append(x)

    return y

print(foo(1))

print(foo(2))


This code defines a function foo that takes two arguments x and y, with y having a default value of an empty list []. Let's break down what happens:

def foo(x, y=[]):: This line defines a function named foo with two parameters, x and y. If y is not provided when calling the function, it defaults to an empty list [].

y.append(x): This line appends the value of x to the list y. Since y is a mutable object and is provided as a default argument, it retains its state across multiple calls to the function.

return y: This line returns the modified list y.

print(foo(1)): This line calls the foo function with x equal to 1. Since y is not provided explicitly, it defaults to [], which becomes [1] after appending 1 to it. So, it prints [1].

print(foo(2)): This line calls the foo function again, this time with x equal to 2. The default value of y is [1] now (the list modified in the previous call). So, 2 is appended to the existing list, resulting in [1, 2]. It prints [1, 2].

However, there's a caveat with this code due to the default mutable argument y=[]. If you call the function foo without providing a value for y, it'll reuse the same list across multiple function calls. This can lead to unexpected behavior if you're not careful. In this case, each time foo is called without specifying y, it keeps appending to the same list object. So, calling foo(1) modifies the list y to [1], and then calling foo(2) appends 2 to the modified list, resulting in [1, 2].

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