Accessing variables ($var and ${var}) in Bash Scripting - Practice Problems & Coding Challenges

Challenge - 5 Problems

💻 Command Output

intermediate

What is the output of this Bash script?

Consider the following Bash script snippet. What will it print when run?

Bash Scripting

name="World"
echo "Hello $name!"
echo "Hello ${name}!"

A Hello ${name}!\nHello World!

B Hello $name!\nHello ${name}!

C Hello World!\nHello $name!

D Hello World!\nHello World!

💡 Hint

Both $var and ${var} are ways to access the value of a variable in Bash.

In Bash, $name and ${name} both access the variable 'name'. So both echo commands print 'Hello World!'.

💻 Command Output

intermediate

What is the output when variable names are adjacent to text?

What will this Bash script print?

Bash Scripting

var="day"
echo "Good $vartoday"
echo "Good ${var}today"

A Good \nGood daytoday

B Good daytoday\nGood daytoday

C Good daytoday\nGood today

D Good \ntoday\nGood today

💡 Hint

Without braces, Bash tries to find a variable named 'vartoday'.

Without braces, $vartoday looks for a variable named 'vartoday' which is empty, so prints 'Good '. With braces, ${var} is 'day', so prints 'Good daytoday'.

📝 Syntax

advanced

Which option correctly accesses a variable followed by a digit?

You have a variable named 'file1'. Which echo command correctly prints its value followed by 'backup'?

Bash Scripting

file1="data.txt"

A echo "${file1backup}"

B echo "$file1backup"

C echo "${file1}backup"

D echo "$file1 backup"

💡 Hint

Use braces to separate variable name from following text.

Option C uses braces to clearly separate the variable 'file1' from the text 'backup'. Option C looks for variable 'file1backup' which is undefined. Option C prints the variable and then a space and 'backup'. Option C looks for variable 'file1backup' which is undefined.

🔧 Debug

advanced

Why does this script print an empty line?

Given the script below, why does the first echo print an empty line?

Bash Scripting

var="test"
echo "$var123"
echo "${var}123"

A Because $var123 looks for variable 'var123' which is not set, so prints empty.

B Because $var123 is a syntax error and stops the script.

C Because $var123 prints the value of 'var' and then '123'.

D Because ${var}123 is invalid syntax.

💡 Hint

Check what variable names Bash tries to expand.

Without braces, $var123 tries to expand variable named 'var123' which is not set, so prints empty. The second echo uses braces to expand 'var' and then adds '123'.

🚀 Application

expert

How many items are in the array after this script runs?

Consider this Bash script. How many elements does the array 'arr' contain after execution?

Bash Scripting

prefix="item"
for i in 1 2 3; do
  arr+=("${prefix}${i}")
done

echo ${#arr[@]}

A Error: arr not declared

B 3

C 1

D 0

💡 Hint

Each loop adds one element to the array.

The loop runs 3 times, each time adding one element to 'arr'. So the array has 3 elements.

Practice

1. In bash scripting, what is the main reason to use ${var} instead of just $var?

easy

A. To clearly separate the variable name from surrounding text

B. To declare a new variable

C. To comment out the variable

D. To convert the variable to uppercase

Solution

1. Step 1: Understand variable access syntax

   In bash, $var accesses the value of a variable named var.

2. Step 2: Purpose of braces

   Using ${var} helps separate the variable name from other text, avoiding confusion when concatenating.

3. Final Answer:

   To clearly separate the variable name from surrounding text -> Option A

4. Quick Check:

   Braces separate variable name from text = A [OK]

Hint: Use braces to avoid mixing variable name with text [OK]

Common Mistakes:

• Thinking braces declare variables
• Using braces to comment variables
• Assuming braces change variable case

2. Which of the following is the correct way to print the value of variable name followed immediately by the string _end in bash?

easy

A. echo $name_end

B. echo $name _end

C. echo ${name_end}

D. echo ${name}_end

Solution

1. Step 1: Understand variable and string concatenation

   To print variable name followed by _end, we must separate the variable name from the string.

2. Step 2: Use braces to separate variable name

   Using ${name}_end correctly prints the variable value plus the string _end. Without braces, $name_end looks for a variable named name_end.

3. Final Answer:

   echo ${name}_end -> Option D

4. Quick Check:

   Braces separate variable name from text = C [OK]

Hint: Use braces when adding text right after variable [OK]

Common Mistakes:

• Forgetting braces and getting wrong variable
• Adding space breaks concatenation
• Using wrong variable name inside braces

3. What will be the output of this bash script?

var="world"
echo "Hello $var!"
echo "Hello ${var}!"
echo "Hello $var123"
echo "Hello ${var}123"

medium

A. Hello world!
Hello world!
Hello world123
Hello 123

B. Hello world!
Hello world!
Hello
Hello world123

C. Hello world!
Hello world!
Hello
Hello 123

D. Hello world!
Hello world!
Hello world123
Hello world123

Solution

1. Step 1: Analyze each echo line

   First two lines print "Hello world!" because $var and ${var} both expand to "world". Third line prints "Hello " because $var123 looks for variable named var123 which is empty. Fourth line prints "Hello world123" because ${var} is "world" plus "123".

2. Step 2: Combine outputs

   Outputs are:
   Hello world!
   Hello world!
   Hello
   Hello world123

3. Final Answer:

   Hello world! Hello world! Hello Hello world123 -> Option B

4. Quick Check:

   Braces separate variable name correctly = D [OK]

Hint: Braces avoid variable name confusion with suffixes [OK]

Common Mistakes:

• Assuming $var123 expands var plus 123
• Ignoring empty variable var123
• Confusing output lines order

4. Given this script snippet:

filename="report"
echo "File: $filename.txt"

Why does it not print File: report.txt as expected?

medium

A. Because echo cannot print variables with dots

B. Because .txt is treated as a command

C. Because $filename.txt looks for variable named filename.txt which is undefined

D. Because variable names cannot contain letters

Solution

1. Step 1: Understand variable expansion with suffix

   Using $filename.txt makes bash look for variable named filename.txt, which does not exist.

2. Step 2: Fix with braces

   Using ${filename}.txt correctly expands filename and appends .txt.

3. Final Answer:

   Because $filename.txt looks for variable named filename.txt which is undefined -> Option C

4. Quick Check:

   Braces needed to separate variable name from suffix = A [OK]

Hint: Use braces when adding dot or special chars after variable [OK]

Common Mistakes:

• Thinking dot is part of variable name
• Assuming echo can't print dots
• Ignoring variable name rules

5. You want to create a bash script that prints the value of variable user followed by the string _backup only if user is not empty. Which is the best way to write the echo command?

hard

A. echo "${user:+${user}_backup}"

B. echo "${user}_backup"

C. echo "${user}_backup" && [ -n "$user" ]

D. echo "$user_backup"

Solution

1. Step 1: Understand conditional variable expansion

   The expression ${user:+${user}_backup} prints ${user}_backup only if user is not empty.

2. Step 2: Compare options

   echo "${user}_backup" always prints _backup even if user is empty. echo "$user_backup" looks for variable user_backup. echo "${user}_backup" && [ -n "$user" ] has wrong syntax mixing echo and condition.

3. Final Answer:

   echo "${user:+${user}_backup}" -> Option A

4. Quick Check:

   Use ${var:+value} to print only if var set = B [OK]

Hint: Use ${var:+value} to print only if var is set [OK]

Common Mistakes:

• Using $user_backup instead of ${user}_backup
• Printing _backup even if user empty
• Mixing echo with condition incorrectly