__name__ and __main__ behavior in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this code when run as a script?

Consider this Python script saved as script.py. What will it print when executed directly with python script.py?

Python

print(f"Top-level __name__ is: {__name__}")

if __name__ == "__main__":
    print("Running as main script")
else:
    print("Imported as a module")

A

Top-level __name__ is: __main__
Running as main script

B

Top-level __name__ is: script
Running as main script

C

Top-level __name__ is: __main__
Imported as a module

D

Top-level __name__ is: script
Imported as a module

💡 Hint

When a Python file is run directly, __name__ is set to "__main__".

When you run a Python file directly, its __name__ variable is set to "__main__". So the first print shows __main__, and the if condition is true, printing "Running as main script".

❓ Predict Output

intermediate

What is the output when this module is imported?

Given the same code below saved as module.py, what will be printed when another script imports it using import module?

Python

print(f"Top-level __name__ is: {__name__}")

if __name__ == "__main__":
    print("Running as main script")
else:
    print("Imported as a module")

A

Top-level __name__ is: __main__
Running as main script

B

Top-level __name__ is: module
Running as main script

C

Top-level __name__ is: module
Imported as a module

D

Top-level __name__ is: __main__
Imported as a module

💡 Hint

When a file is imported, __name__ is the module's name, not "__main__".

When imported, __name__ is the module's filename without .py, here "module". So the first print shows "module" and the else branch runs, printing "Imported as a module".

🔧 Debug

advanced

Why does this code print nothing when run directly?

This code is saved as test.py. When run with python test.py, it prints nothing. Why?

Python

def main():
    print("Hello from main")

if __name__ == "__main__":
    main

A Because __name__ is not "__main__" when run directly.

B Because main is referenced but not called (missing parentheses).

C Because print statement is inside a function that is never defined.

D Because the code has a syntax error and does not run.

💡 Hint

Check if the function main is actually executed.

The code references the function main but does not call it. It should be main() with parentheses to run the function and print the message.

📝 Syntax

advanced

Which option causes a SyntaxError?

Which of these code snippets will cause a SyntaxError when run?

A

if __name__ = "__main__":
    print("Running")

B

if __name__ == "__main__":
    print("Running")

C

)"gninnuR"(tnirp    
:"__niam__" == __eman__ fi

D

if __name__ == "__main__":
    print("Running")
else:
    print("Not main")

💡 Hint

Check the operator used in the if condition.

Option A uses a single equals sign (=) instead of double equals (==) in the if condition, which is invalid syntax in Python.

🚀 Application

expert

What is the output of this code when run directly?

Analyze this code saved as app.py. What will it print when executed with python app.py?

Python

def greet():
    print(f"Hello from {__name__}!")

if __name__ == "__main__":
    greet()
else:
    print(f"Imported module: {__name__}")

A No output

B Imported module: app

C Hello from app!

D Hello from __main__!

💡 Hint

Remember what __name__ is when running a script directly.

When run directly, __name__ is "__main__", so greet() prints "Hello from __main__!".

Practice

1. What does the special variable __name__ contain when a Python file is run directly?

easy

A. The file's directory path

B. The file's extension

C. "__main__"

D. The Python version

Solution

1. Step 1: Understand the role of __name__

   When a Python file runs directly, __name__ is set to the string "__main__".

2. Step 2: Differentiate direct run vs import

   If the file is imported, __name__ is the module's name, not "__main__".

3. Final Answer:

   "__main__" -> Option C

4. Quick Check:

   __name__ = "__main__" when run directly [OK]

Hint: Direct run sets __name__ to "__main__" [OK]

Common Mistakes:

• Thinking __name__ holds file path
• Confusing __name__ with Python version
• Assuming __name__ is always module name

2. Which of the following is the correct syntax to run code only when a Python file is executed directly?

easy

A. if __name__ == '__main__':

B. if __main__ == '__name__':

C. if main == '__name__':

D. if __name__ = '__main__':

Solution

1. Step 1: Check correct variable and string

   The variable is __name__ and the string to compare is "__main__".

2. Step 2: Verify syntax correctness

   Use double equals == for comparison, and colons to start the block.

3. Final Answer:

   if __name__ == '__main__': -> Option A

4. Quick Check:

   Correct syntax uses == and exact names [OK]

Hint: Use if __name__ == '__main__': exactly [OK]

Common Mistakes:

• Using single equals (=) instead of double (==)
• Swapping __name__ and __main__
• Missing colon at end of if statement

3. What will be the output when running this Python file directly?

def greet():
    print('Hello!')

if __name__ == '__main__':
    greet()

medium

A. Hello! Hello!

B. No output

C. Error: greet() undefined

D. Hello!

Solution

1. Step 1: Check if condition when run directly

   Since the file runs directly, __name__ == '__main__' is True, so greet() is called.

2. Step 2: Understand greet() function output

   The function prints "Hello!" once.

3. Final Answer:

   Hello! -> Option D

4. Quick Check:

   Function called once prints "Hello!" [OK]

Hint: Direct run triggers greet() printing once [OK]

Common Mistakes:

• Thinking greet() runs twice
• Assuming no output without main guard
• Confusing function call with definition

4. Identify the error in this code snippet:

def main():
    print('Running main')

if __name__ = '__main__':
    main()

medium

A. IndentationError in function definition

B. SyntaxError due to single equals (=) in if condition

C. NameError because main() is undefined

D. No error, code runs fine

Solution

1. Step 1: Check the if condition syntax

   The condition uses single equals (=) which is assignment, not comparison, causing SyntaxError.

2. Step 2: Confirm function and indentation

   The function main() is defined correctly and indentation is fine.

3. Final Answer:

   SyntaxError due to single equals (=) in if condition -> Option B

4. Quick Check:

   Use == for comparison, not = [OK]

Hint: Use == in if condition, not = [OK]

Common Mistakes:

• Using = instead of ==
• Assuming function undefined error
• Ignoring indentation correctness

5. You have two Python files:

# file1.py
def greet():
    print('Hi from file1')

if __name__ == '__main__':
    greet()

# file2.py
import file1
print('In file2')

What will be the output when you run file2.py?

hard

A. In file2

B. No output

C. Hi from file1

D. Hi from file1
In file2

Solution

1. Step 1: Understand import behavior with __name__

   When file1 is imported, its __name__ is "file1", not "__main__", so greet() inside the if block does NOT run.

2. Step 2: Check what runs in file2.py

   Only print('In file2') runs, so output is just "In file2".

3. Final Answer:

   In file2 -> Option A

4. Quick Check:

   Import skips main block, prints only file2 message [OK]

Hint: Imported file skips if __name__ == '__main__' block [OK]

Common Mistakes:

• Assuming greet() runs on import
• Expecting both prints always
• Confusing __name__ values on import