Sunday 26 May 2024

Python Coding challenge - Day 217 | What is the output of the following Python Code?

 


The function foo is defined with a default argument x which is an empty list []. Here is the breakdown of the function and the output:

Function Definition

def foo(x=[]):

    x.append(1)

    return x

Key Points

  • Default Argument: The default value of x is [], an empty list.
  • Appending to List: Inside the function, 1 is appended to the list x.
  • Returning the List: The modified list x is returned.

Default Mutable Arguments

In Python, default arguments are evaluated only once when the function is defined, not each time the function is called. This means that if the default argument is a mutable object like a list, and you modify this object, the changes will persist across function calls.

Execution

First Call:

print(foo())

foo is called without any arguments, so the default empty list [] is used.

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

The list [1] is returned and printed.

Second Call:

print(foo())

foo is called again without any arguments.

This time, the same list (now [1] from the first call) is used as the default value.

1 is appended again, making the list [1, 1].

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

Output

Here is the complete code and its output:

def foo(x=[]):

    x.append(1)

    return x

print(foo())  # Output: [1]

print(foo())  # Output: [1, 1]

Explanation

First print(foo()):

The default list x is [].

1 is appended to x, resulting in [1].

[1] is printed.

Second print(foo()):

The default list x is now [1] (the same list used in the previous call).

1 is appended again, resulting in [1, 1].

[1, 1] is printed.

Conclusion

The key takeaway is that the default argument x=[] is not re-evaluated each time foo is called. Instead, the same list object is used across all calls to the function, leading to the list accumulating all the appended 1s over multiple calls.

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