Python Coding challenge - Day 1083| What is the output of the following Python Code?

 

Code Explanation:

🔹 1️⃣ Defining Descriptor Class D

class D:

A class named D is created.

This class will act as a descriptor.

🔹 2️⃣ Defining the __get__ Method

def __get__(self, obj, objtype):

    return 100

This method runs when the attribute is accessed.

Parameters:

self → descriptor object

obj → instance accessing the attribute (a)

objtype → class of the instance (A)

Behavior:

return 100

So whenever the attribute is accessed, the value 100 is returned.

🔹 3️⃣ Defining the __set__ Method

def __set__(self, obj, value):

    obj.__dict__['x'] = value

This method runs when the attribute is assigned.

Example:

a.x = 5

Execution:

obj.__dict__['x'] = value

So internally Python stores:

a.__dict__['x'] = 5

🔹 4️⃣ Defining Class A

class A:

A class named A is created.

🔹 5️⃣ Assigning Descriptor to Class Attribute

x = D()

Here:

An instance of D is assigned to x.

Internally:

A.x → descriptor object

So x is now controlled by the descriptor D.

🔹 6️⃣ Creating an Object

a = A()

An instance a of class A is created.

Initially:

a.__dict__ = {}

🔹 7️⃣ Assigning Value to a.x

a.x = 5

Python sees that x is a descriptor with __set__.

So it calls:

D.__set__(descriptor, a, 5)

Inside the method:

a.__dict__['x'] = 5

Now:

a.__dict__ = {'x': 5}

🔹 8️⃣ Accessing a.x

print(a.x)

Now Python performs attribute lookup.

Lookup order:

1️⃣ Data descriptor

Since x is a data descriptor, Python calls:

D.__get__(descriptor, a, A)

Inside __get__:

return 100

🔹 9️⃣ Instance Dictionary Ignored

Even though:

a.__dict__['x'] = 5

Python ignores it because:

⚠ Data descriptors take priority over instance attributes.

✅ Final Output

100

400 Days Python Coding Challenges with Explanation