Python Coding challenge - Day 1127| What is the output of the following Python Code?

 

Code Explanation:

🔹 1. Class Definition

class Test:

    x = []

✅ Explanation:

A class named Test is created.

x = [] is a class variable (shared by all objects).

This means:

Only one list exists in memory.

All instances (a, b, etc.) will refer to the same list unless overridden.

🔹 2. Constructor (__init__ method)

def __init__(self, value):

    self.x.append(value)

✅ Explanation:

This method runs whenever an object is created.

self refers to the current object.

self.x.append(value):

Python first looks for x inside the instance.

Not found → it looks in the class.

Finds x (the shared list).

So, it appends the value to the same shared list.

🔹 3. Creating First Object

a = Test(1)

✅ What happens:

Object a is created.

__init__(1) runs.

self.x.append(1) → list becomes:

[1]

🔹 4. Creating Second Object

b = Test(2)

✅ What happens:

Object b is created.

__init__(2) runs.

Again, self.x refers to the same class list.

2 is appended → list becomes:

[1, 2]

🔹 5. Printing Values

print(a.x, b.x)

✅ Explanation:

Both a.x and b.x refer to the same list.

So output is:

[1, 2] [1, 2]

⚠️ Key Concept (Very Important)

🔸 Class Variable vs Instance Variable

Type Defined Where Shared?

Class Variable Inside class ✅ Yes

Instance Variable Inside __init__ using self ❌ No

🔥 Why This Happens

Because:

x = []

is defined at class level, not inside __init__.

✅ How to Fix (If You Want Separate Lists)

class Test:

    def __init__(self, value):

        self.x = []      # instance variable

        self.x.append(value)

✔️ Output now:

[1] [2]

🎯 Final Answer

[1, 2] [1, 2]

 Book:  500 Days Python Coding Challenges with Explanation