Tuesday 26 March 2024

Python Coding challenge - Day 158 | What is the output of the following Python Code?

 

This code defines a function foo that takes a single argument x. The argument x is initialized with a default value of an empty list [].

def foo(x=[]):

    x.append(1)

    return x

Here's what happens when you call foo() multiple times:

First Call (print(foo())):

foo() is called without any argument, so x defaults to an empty list [].

Inside the function, 1 is appended to the list x, modifying it to [1].

The modified list [1] is returned and printed.

Second Call (print(foo())):


Since the default argument x retains its value between calls, it still holds the modified list [1] from the previous call.

1 is appended to the existing list x, resulting in [1, 1].

The modified list [1, 1] is returned and printed.

Third Call (print(foo())):


Similar to the second call, the default argument x still holds the modified list [1, 1].

Another 1 is appended to the list x, making it [1, 1, 1].

The modified list [1, 1, 1] is returned and printed.

So, the output of the three function calls will be:

[1]

[1, 1]

[1, 1, 1]

It's important to note that the default argument x=[] is evaluated only once when the function foo is defined. This means that every time you call foo() without passing an argument explicitly, the same list object (which was created when the function was defined) is used. This can lead to unexpected behavior if you're not careful, especially when dealing with mutable default arguments like lists or dictionaries.

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