Python Coding challenge - Day 439| What is the output of the following Python Code?

 

Step-by-step Explanation:

Initialize an empty list funcs:

funcs = []

This is just creating an empty list, where functions will be stored.

Loop over the range 3:

for i in range(3):

The for loop runs three times, with i taking values 0, 1, and 2 in each iteration.

Define the function f() inside the loop:

def f(): return i

A function f() is defined within the loop that returns the value of i when called.

At this point, the function f is being defined, but it doesn't immediately execute. The definition of f happens in the current scope of the loop, which means that f will "remember" the variable i when it is called, but it does not capture the value of i at the time the function was defined (which is crucial to understanding the behavior).

Append the function f() to funcs list:

funcs.append(f)

The function f (which is defined within the loop) is added to the funcs list. However, because of the late binding behavior in Python, the function f does not capture the value of i at the moment it is defined. Instead, it captures the reference to i, meaning it always uses the current value of i when it is called.

This behavior will be important later: after the loop finishes, all functions in funcs will reference the final value of i, which is 2 (since the loop ends when i = 2).

Calling each function in the funcs list:

print([fn() for fn in funcs])

This line creates a list comprehension that calls each function (fn()) stored in the funcs list and prints the result. Let’s analyze what happens when each function is called:

When calling any function in funcs, the value of i is not the value i at the time the function was added to the list; instead, it is the final value of i after the loop ends.

Since the loop finishes with i = 2, all functions in funcs will return the value 2.

Therefore, when each of the functions in funcs is called, they all return 2.

Output:

[2, 2, 2]