Why scope matters in Python - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print the value of the variable inside the function.

Python

def greet():
    message = 'Hello!'
    print([1])
greet()

Drag options to blanks, or click blank then click option'

Aprint

Bgreet

Cmessage

DHello

Attempts:

3 left

💡 Hint

Common Mistakes

Using the function name instead of the variable.

Trying to print a string without quotes.

Using a variable not defined in the function.

The variable message is defined inside the function, so we use its name to print its value.

2fill in blank

medium

Complete the code to access the global variable inside the function.

Python

count = 5
def show_count():
    print([1])
show_count()

Drag options to blanks, or click blank then click option'

Acount

B5

Cshow_count

Dprint

Attempts:

3 left

💡 Hint

Common Mistakes

Using the function name instead of the variable.

Trying to print the number directly without the variable.

Using a variable not defined globally.

The variable count is defined outside the function, but it can be accessed inside because it is global.

3fill in blank

hard

Fix the error by completing the code to modify the global variable inside the function.

Python

counter = 10
def increase():
    global [1]
    counter += 1
increase()
print(counter)

Drag options to blanks, or click blank then click option'

Acounter

Bincrease

Ccount

Dglobal

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different variable name in the global statement.

Forgetting the global keyword.

Trying to modify the variable without declaring it global.

To change a global variable inside a function, you must declare it with global and use the exact variable name.

4fill in blank

hard

Fill both blanks to create a local variable and print it inside the function.

Python

def show():
    [1] = 'Local'
    print([2])
show()

Drag options to blanks, or click blank then click option'

Atext

Bmessage

Dprint

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names for assignment and print.

Trying to print a function name.

Using a keyword like print as a variable.

The variable text is created locally and printed inside the function.

5fill in blank

hard

Fill all three blanks to create a global variable, modify it inside a function, and print the updated value.

Python

value = 3
def update():
    global [1]
    [2] += 7
update()
print([3])

Drag options to blanks, or click blank then click option'

Avalue

Dupdate

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in different places.

Forgetting the global keyword.

Trying to print the function name instead of the variable.

We declare value as global, modify it inside the function, and print it outside.

Practice

1. What does the term scope mean in Python programming?

easy

A. The area where a variable can be accessed or used

B. The size of a variable in memory

C. The speed at which a program runs

D. The type of a variable

Solution

1. Step 1: Understand variable accessibility

   Scope defines where a variable can be accessed in the code.

2. Step 2: Differentiate scope from other concepts

   Scope is not about size, speed, or type but about accessibility.

3. Final Answer:

   The area where a variable can be accessed or used -> Option A

4. Quick Check:

   Scope = variable accessibility [OK]

Hint: Scope means where variables can be used in code [OK]

Common Mistakes:

• Confusing scope with variable size
• Thinking scope affects program speed
• Mixing scope with variable type

2. Which of the following is the correct way to declare a global variable inside a function?

easy

A. global = x

B. def global x

C. global x

D. var global x

Solution

1. Step 1: Recall Python syntax for global variables

   To modify a global variable inside a function, use the keyword global followed by the variable name.

2. Step 2: Check each option's syntax

   Only global x is valid Python syntax; others are incorrect.

3. Final Answer:

   global x -> Option C

4. Quick Check:

   Use 'global' keyword correctly [OK]

Hint: Use 'global' keyword before variable name inside functions [OK]

Common Mistakes:

• Using 'def' or 'var' with global
• Assigning 'global = x' which is invalid
• Forgetting to declare global before use

3. What will be the output of this code?

count = 5

def increment():
    count = 10
    print(count)

increment()
print(count)

medium

A. 10\n5

B. 10\n10

C. 5\n5

D. 5\n10

Solution

1. Step 1: Analyze variable scope inside the function

   Inside increment(), count = 10 creates a local variable named count that shadows the global one.

2. Step 2: Check print statements

   The first print inside the function prints local count (10). The second print outside prints global count (5).

3. Final Answer:

   10 5 -> Option A

4. Quick Check:

   Local shadows global inside function [OK]

Hint: Local variables inside functions don't change globals unless declared [OK]

Common Mistakes:

• Assuming global variable changes inside function without 'global'
• Confusing which 'count' is printed
• Expecting both prints to show 10

4. Find the error in this code related to variable scope:

def add_one():
    x += 1
    print(x)

x = 5
add_one()

medium

A. No error, output will be 6

B. SyntaxError due to missing colon

C. NameError because x is not defined anywhere

D. UnboundLocalError because x is used before assignment inside function

Solution

1. Step 1: Understand variable modification inside function

   Inside add_one(), x += 1 tries to modify x locally, but x is not declared local or global.

2. Step 2: Identify error type

   Python raises UnboundLocalError because it thinks x is local but it's used before assignment.

3. Final Answer:

   UnboundLocalError because x is used before assignment inside function -> Option D

4. Quick Check:

   Modifying global without 'global' causes UnboundLocalError [OK]

Hint: Declare 'global x' to modify global variable inside function [OK]

Common Mistakes:

• Thinking it's a NameError
• Expecting code to print 6 without error
• Ignoring need for 'global' keyword

5. Given this code, what will be the output?

def outer():
    x = 'local'
    def inner():
        nonlocal x
        x = 'nonlocal'
        print('inner:', x)
    inner()
    print('outer:', x)

outer()

hard

A. inner: local\nouter: local

B. inner: nonlocal\nouter: nonlocal

C. inner: nonlocal\nouter: local

D. SyntaxError due to nonlocal usage

Solution

1. Step 1: Understand 'nonlocal' keyword effect

   The nonlocal keyword allows inner() to modify x defined in outer(), not create a new local variable.

2. Step 2: Trace print outputs

   inner() prints inner: nonlocal after changing x. Then outer() prints outer: nonlocal showing the updated value.

3. Final Answer:

   inner: nonlocal outer: nonlocal -> Option B

4. Quick Check:

   'nonlocal' changes outer function variable [OK]

Hint: Use 'nonlocal' to modify outer function variables inside nested functions [OK]

Common Mistakes:

• Thinking 'nonlocal' causes syntax error
• Assuming inner creates a new local variable
• Expecting outer's x to remain 'local'