Local variables (local keyword) in Bash Scripting - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a local variable inside the function.

Bash Scripting

my_function() {
  [1] my_var="Hello"
  echo "$my_var"
}

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

Alocal

Bglobal

Cdeclare

Dvar

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'global' instead of 'local' which is not valid in bash.

Using 'var' which is not a bash keyword.

Using 'declare' without the local keyword does not limit scope.

The 'local' keyword declares a variable that is only visible inside the function.

2fill in blank

medium

Complete the code to print the local variable inside the function.

Bash Scripting

function greet() {
  local message="Hi"
  echo [1]
}
greet

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

Amessage

B$message

C"message"

D'message'

Attempts:

3 left

💡 Hint

Common Mistakes

Printing the variable name without $ which outputs the name, not the value.

Using quotes around the variable name which prints the literal string.

To print the value of a variable in bash, prefix it with a dollar sign ($).

3fill in blank

hard

Fix the error in the function by declaring the variable local.

Bash Scripting

count=5
increment() {
  [1] count=$((count + 1))
  echo $count
}
increment
 echo $count

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

Aexport

Bglobal

Clocal

Ddeclare

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'global' which is not a bash keyword.

Using 'export' which affects environment variables, not local scope.

Declaring 'count' as local inside the function prevents changing the global variable.

4fill in blank

hard

Fill both blanks to declare a local variable and assign it a value.

Bash Scripting

my_func() {
  [1] [2]="World"
  echo "$greeting"
}
my_func

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

Alocal

Bgreeting

Cmessage

Dvar

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'var' which is not a bash keyword.

Using a variable name different from the one echoed.

Use 'local' to declare the variable and 'greeting' as the variable name.

5fill in blank

hard

Fill all three blanks to declare a local variable, assign it, and print it.

Bash Scripting

function say() {
  [1] [2]="Hello"
  echo [3]
}
say

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

Alocal

Bgreet

C$greet

Dmessage

Attempts:

3 left

💡 Hint

Common Mistakes

Not using local keyword.

Printing variable name without $.

Using different variable names inconsistently.

Declare 'greet' as local, assign it, and print its value with $.

Practice

1. What does the local keyword do in a Bash function?

easy

A. It deletes the variable after the function ends.

B. It makes the variable global and accessible everywhere.

C. It exports the variable to child processes.

D. It creates a variable that exists only inside the function.

Solution

1. Step 1: Understand variable scope in Bash functions

   Variables declared with local exist only inside the function where they are declared.

2. Step 2: Compare with other variable types

   Global variables exist outside functions; local prevents changes outside the function.

3. Final Answer:

   It creates a variable that exists only inside the function. -> Option D

4. Quick Check:

   Local variables = function-only scope [OK]

Hint: Local means variable stays inside the function only [OK]

Common Mistakes:

• Thinking local variables are global
• Confusing local with export
• Assuming local deletes variables

2. Which of the following is the correct way to declare a local variable named count inside a Bash function?

easy

A. count local=10

B. var local count=10

C. local count=10

D. declare count local=10

Solution

1. Step 1: Recall correct syntax for local variables

   The correct syntax is local variable_name=value.

2. Step 2: Check each option

   Only local count=10 matches the correct syntax.

3. Final Answer:

   local count=10 -> Option C

4. Quick Check:

   local + variable = correct syntax [OK]

Hint: Use 'local var=value' inside functions [OK]

Common Mistakes:

• Placing 'local' after variable name
• Using 'var' keyword which doesn't exist
• Mixing 'declare' with 'local' incorrectly

3. What will be the output of this script?

#!/bin/bash
value=5
function test_func() {
  local value=10
  echo $value
}
test_func
echo $value

medium

A. 10
5

B. 5
10

C. 10
10

D. 5
5

Solution

1. Step 1: Analyze local variable inside function

   Inside test_func, value is local and set to 10, so echo $value prints 10.

2. Step 2: Check global variable outside function

   Outside the function, value remains 5 because local variable does not affect it.

3. Final Answer:

   10 5 -> Option A

4. Quick Check:

   Local inside function, global outside = 10 then 5 [OK]

Hint: Local variable hides global only inside function [OK]

Common Mistakes:

• Assuming local changes global variable
• Expecting both outputs to be same
• Confusing order of printed values

4. Identify the error in this Bash function:

function example() {
  local=5
  echo $local
}

medium

A. Using 'local' as a variable name causes a syntax error.

B. Missing 'local' keyword before variable name.

C. Function missing parentheses after name.

D. No error; script runs fine.

Solution

1. Step 1: Check variable naming rules

   Using local as a variable name conflicts with the reserved keyword local.

2. Step 2: Understand syntax error cause

   Assigning local=5 tries to overwrite the keyword, causing a syntax error.

3. Final Answer:

   Using 'local' as a variable name causes a syntax error. -> Option A

4. Quick Check:

   Reserved keywords can't be variable names [OK]

Hint: Don't name variables 'local'--it's a reserved keyword [OK]

Common Mistakes:

• Thinking 'local' can be a variable name
• Ignoring syntax errors from reserved words
• Missing parentheses is not the error here

5. You want to write a Bash function that increments a counter variable without affecting the global variable of the same name. Which approach correctly uses local to achieve this?

hard

A. Increment the global variable directly without declaring local.

B. Declare local counter inside the function and increment it there.

C. Use export counter inside the function before incrementing.

D. Declare counter outside the function and use local globally.

Solution

1. Step 1: Understand local variable purpose

   Declaring local counter inside the function creates a separate variable that does not affect the global one.

2. Step 2: Check other options

   Incrementing global directly changes global variable; export shares variables with child processes; declaring local outside function is invalid.

3. Final Answer:

   Declare local counter inside the function and increment it there. -> Option B

4. Quick Check:

   Local inside function isolates variable changes [OK]

Hint: Use 'local' inside function to isolate variable changes [OK]

Common Mistakes:

• Modifying global variable unintentionally
• Misusing export for local scope
• Declaring local variables outside functions