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.