Double quotes (variable expansion) 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:

name=World

echo "Hello $name!"
echo 'Hello $name!'

What will be printed when this script runs?

Bash Scripting

name=World

echo "Hello $name!"
echo 'Hello $name!'

A Hello World!\nHello $name!

B Hello $name!\nHello $name!

C Hello World!\nHello World!

D Hello $name!\nHello World!

💡 Hint

Remember that double quotes allow variable expansion, but single quotes do not.

In Bash, variables inside double quotes are replaced by their values. Inside single quotes, variables are treated as literal text.

💻 Command Output

intermediate

What does this script print?

Given this Bash script:

var=5

echo "$var is a number"
echo '$var is a number'

What is the output?

Bash Scripting

var=5

echo "$var is a number"
echo '$var is a number'

A $var is a number\n$var is a number

B $var is a number\n5 is a number

C 5 is a number\n$var is a number

D 5 is a number\n5 is a number

💡 Hint

Check how variables behave inside double vs single quotes.

Double quotes expand variables, so $var becomes 5. Single quotes print the text as-is.

🔧 Debug

advanced

Why does this script not print the variable value?

Look at this Bash script:

greeting=Hello

echo '$greeting, friend!'

Why does it print '$greeting, friend!' instead of 'Hello, friend!'?

Bash Scripting

greeting=Hello

echo '$greeting, friend!'

A Because the variable greeting is not defined properly.

B Because single quotes prevent variable expansion, so $greeting is printed literally.

C Because echo does not support variables inside quotes.

D Because the variable name is misspelled.

💡 Hint

Try changing single quotes to double quotes.

Single quotes in Bash treat everything inside as literal text, so variables are not expanded.

💻 Command Output

advanced

What is the output of this script with mixed quotes?

Consider this Bash script:

user=Alice

echo "User is '$user'"
echo 'User is "$user"'

What will be printed?

Bash Scripting

user=Alice

echo "User is '$user'"
echo 'User is "$user"'

A User is 'Alice'\nUser is "$user"

B User is '$user'\nUser is "Alice"

C User is 'Alice'\nUser is "Alice"

D User is $user\nUser is $user

💡 Hint

Remember how quotes inside quotes affect variable expansion.

Inside double quotes, single quotes are literal, so $user expands. Inside single quotes, double quotes and variables are literal.

🚀 Application

expert

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

Given this Bash script:

items="apple banana cherry"

arr=($items)

echo ${#arr[@]}

What number will be printed?

Bash Scripting

items="apple banana cherry"

arr=($items)

echo ${#arr[@]}

A 1

B 0

C SyntaxError

D 3

💡 Hint

Check how the string is split into array elements.

The string is split by spaces into three elements: apple, banana, cherry. So the array length is 3.

Practice

1. What happens when you use double quotes around a variable in bash, like "$name"?

easy

A. The variable name is replaced by its value inside the quotes.

B. The variable name is treated as plain text, not replaced.

C. The shell throws a syntax error.

D. The variable is removed from the string.

Solution

1. Step 1: Understand variable expansion in double quotes

   In bash, variables inside double quotes are replaced by their values.

2. Step 2: Compare with other quote types

   Single quotes prevent expansion, but double quotes allow it while preserving spaces.

3. Final Answer:

   The variable name is replaced by its value inside the quotes. -> Option A

4. Quick Check:

   Double quotes expand variables = A [OK]

Hint: Double quotes expand variables, single quotes do not [OK]

Common Mistakes:

• Thinking variables inside double quotes are not expanded
• Confusing single and double quotes
• Expecting syntax error with double quotes

2. Which of the following is the correct way to print the value of variable user with a greeting using double quotes in bash?

easy

A. echo 'Hello, $user!'

B. echo "Hello, 'user'!"

C. echo Hello, $user!

D. echo "Hello, $user!"

Solution

1. Step 1: Identify correct quoting for variable expansion

   Double quotes allow variable $user to expand inside the string.

2. Step 2: Check each option

   echo 'Hello, $user!' uses single quotes, so $user is literal. echo Hello, $user! lacks quotes, which can cause word splitting. echo "Hello, 'user'!" quotes 'user' literally inside double quotes.

3. Final Answer:

   echo "Hello, $user!" -> Option D

4. Quick Check:

   Use double quotes to expand variables in strings = B [OK]

Hint: Use double quotes to expand variables inside strings [OK]

Common Mistakes:

• Using single quotes which prevent expansion
• Not quoting strings with spaces
• Mixing single and double quotes incorrectly

3. Given the script:

name="Alice"
echo "Hello, $name!"
echo 'Hello, $name!'

What is the output?

medium

A. Hello, Alice!
Hello, Alice!

B. Hello, Alice!
Hello, $name!

C. Hello, $name!
Hello, Alice!

D. Hello, $name!
Hello, $name!

Solution

1. Step 1: Analyze first echo with double quotes

   Double quotes expand $name to "Alice", so output is "Hello, Alice!".

2. Step 2: Analyze second echo with single quotes

   Single quotes prevent expansion, so output is literally "Hello, $name!".

3. Final Answer:

   Hello, Alice! Hello, $name! -> Option B

4. Quick Check:

   Double quotes expand, single quotes do not = A [OK]

Hint: Double quotes expand variables; single quotes print literally [OK]

Common Mistakes:

• Assuming single quotes expand variables
• Confusing output order
• Ignoring newline between echo commands

4. Consider this script snippet:

file="my file.txt"
echo $file
ls "$file"

What is the problem and how to fix it?

medium

A. First echo should use double quotes to preserve spaces: echo "$file".

B. Second line should remove quotes: ls $file.

C. No problem; script works fine.

D. Variable should be declared without quotes: file=my file.txt.

Solution

1. Step 1: Identify issue with first echo

   Without quotes, echo $file splits "my file.txt" into two words, printing as two separate arguments.

2. Step 2: Fix by adding double quotes

   Using echo "$file" preserves the space and prints the full filename correctly.

3. Final Answer:

   First echo should use double quotes to preserve spaces: echo "$file". -> Option A

4. Quick Check:

   Quote variables to keep spaces intact = C [OK]

Hint: Quote variables to keep spaces safe in commands [OK]

Common Mistakes:

• Not quoting variables with spaces
• Removing quotes from commands that need them
• Declaring variables without quotes causing errors

5. You want to create a script that prints a message with a variable count inside double quotes, but only if count is not empty. Which is the best way to write this?

hard

A. if [ $count ]; then echo "Count is $count"; fi

B. if [ $count ]; then echo 'Count is $count'; fi

C. if [ "$count" ]; then echo "Count is $count"; fi

D. if [ "$count" ]; then echo 'Count is $count'; fi

Solution

1. Step 1: Check condition with quotes

   Using [ "$count" ] safely tests if count is non-empty, even if it contains spaces or is unset.

2. Step 2: Print message with double quotes

   Using double quotes in echo expands $count properly inside the string.

3. Final Answer:

   if [ "$count" ]; then echo "Count is $count"; fi -> Option C

4. Quick Check:

   Quote variables in tests and output for safety = D [OK]

Hint: Quote variables in tests and output to avoid errors [OK]

Common Mistakes:

• Not quoting variable in test causing errors
• Using single quotes in echo preventing expansion
• Testing variable without quotes causing syntax errors