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.