Python Coding Challange - Question With Answer(01210325)

 

Key Points:

1. Local Scope of name

   • The variable name is defined inside the function set_name().
   • This means name exists only within the function's scope and cannot be accessed outside.

2. Function Execution (set_name())

   • The function runs and assigns 'Python' to name, but since name is local, it is lost after the function finishes execution.

3. Error in print(name)

   • The print(name) statement is outside the function and cannot access name because name only exists inside set_name().
   • Result: Python will throw a NameError: name 'name' is not defined.

How to Fix It?

If you want name to be accessible outside the function, you can:

Solution 1: Return the value

def set_name():
    return 'Python'

name = set_name() # Store the returned value in a variable
print(name) # Output: Python

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Solution 2: Use a Global Variable (Not Recommended)

def set_name():
    global name # Declare 'name' as global
    name = 'Python' 
set_name() 
print(name) # Output: Python

• This works but using global is not recommended unless absolutely necessary, as it can lead to unexpected behavior.
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