Default values (${var:-default}) in Bash Scripting - Practice Problems & Coding Challenges

Challenge - 5 Problems

💻 Command Output

intermediate

What is the output of this script using default values?

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

Bash Scripting

unset NAME
echo "Hello, ${NAME:-Guest}!"

A Hello, ${NAME:-Guest}!

B Hello, !

C Hello, Guest!

D Hello, NAME!

💡 Hint

If a variable is not set, the default value after :- is used.

Since NAME is unset, ${NAME:-Guest} uses the default 'Guest'. So the output is 'Hello, Guest!'.

💻 Command Output

intermediate

Output when variable is empty string with default value

What will this script print?

Bash Scripting

NAME=""
echo "Welcome, ${NAME:-User}!"

A Welcome, ${NAME:-User}!

B Welcome, User!

C Welcome, !

D Welcome, ""!

💡 Hint

Default value is used only if variable is unset or null (empty).

The variable NAME is set but empty (null). The :- operator uses the default 'User' when the variable is unset or null. So the output is 'Welcome, User!'.

💻 Command Output

advanced

Difference between ${var:-default} and ${var-default}

Given this script, what is the output?

Bash Scripting

NAME=""
echo "A: ${NAME:-Guest}"
echo "B: ${NAME-Guest}"

A

A: Guest
B: 

B

A: 
B: Guest

C

A: Guest
B: Guest

D

A: 
B: 

💡 Hint

:- uses default if variable is unset or null; - uses default only if unset.

NAME is set but empty. ${NAME:-Guest} treats empty as null and uses default 'Guest'. ${NAME-Guest} treats empty as set and uses empty string. So output is A: Guest and B: (empty).

🔧 Debug

advanced

Why does this script not print the default value?

This script is intended to print 'Hello, Guest!' if NAME is unset or empty. Why does it print 'Hello, !' instead? CODE: NAME="" echo "Hello, ${NAME-default}!"

A Because the default value must be in quotes to work.

B Because the syntax ${NAME-default} is invalid and causes empty output.

C Because the variable NAME is not exported to the environment.

D Because ${NAME-default} uses default only if NAME is unset, not if empty.

💡 Hint

Check difference between - and :- in parameter expansion.

The operator - uses the default only if the variable is unset. Since NAME is set (empty string), default is not used, so output is empty.

🚀 Application

expert

Write a script snippet that prints a default value only if variable is unset or empty

Which option correctly prints the value of VAR if set and not empty, or 'default_value' otherwise?

A echo "${VAR:-default_value}"

B echo "${VAR:+default_value}"

C echo "${VAR:=default_value}"

D echo "${VAR-default_value}"

💡 Hint

Use the operator that substitutes default if variable is unset or empty.

The :- operator substitutes the default if VAR is unset or empty. - substitutes default only if unset. := assigns default to VAR if unset or empty. + substitutes default only if VAR is set and not empty.

Practice

1. What does the expression ${var:-default} do in a bash script?

easy

A. It uses the value of var if set; otherwise, it uses default without changing var.

B. It sets var to default if var is unset or empty.

C. It always replaces var with default.

D. It throws an error if var is unset or empty.

Solution

1. Step 1: Understand the syntax of ${var:-default}

   This syntax means: use var if it is set and not empty; otherwise, use default.

2. Step 2: Check if var is changed

   This expression does not assign or change var, it only provides a fallback value for use.

3. Final Answer:

   It uses the value of var if set; otherwise, it uses default without changing var. -> Option A

4. Quick Check:

   Default value used only if var is empty or unset [OK]

Hint: Remember: :- means use default if empty or unset [OK]

Common Mistakes:

• Thinking it changes the variable's value
• Confusing :- with := which assigns default
• Assuming it throws an error if unset

2. Which of the following is the correct syntax to print a variable name or use "Guest" if name is empty or unset?

easy

A. echo ${name=Guest}

B. echo ${name?Guest}

C. echo ${name+Guest}

D. echo ${name:-Guest}

Solution

1. Step 1: Identify the correct default value syntax

   The syntax ${var:-default} prints var if set; otherwise, prints default.

2. Step 2: Check each option

   echo ${name=Guest} assigns default if unset, not just prints. echo ${name+Guest} prints Guest only if name is set. echo ${name?Guest} throws error if name is unset. Only echo ${name:-Guest} prints name or "Guest" safely.

3. Final Answer:

   echo ${name:-Guest} -> Option D

4. Quick Check:

   Use :- for default fallback [OK]

Hint: Use :- to print default without assignment [OK]

Common Mistakes:

• Using = which assigns instead of fallback
• Using + which prints default only if var is set
• Using ? which causes error if var unset

3. What is the output of this script?

unset user
user=""
echo "User: ${user:-anonymous}"

medium

A. User: anonymous

B. User:

C. User: user

D. User: ${user:-anonymous}

Solution

1. Step 1: Analyze variable state

   The variable user is first unset, then set to an empty string.

2. Step 2: Apply default value logic

   Since user is empty (empty string counts as unset for :-), the expression uses the default "anonymous".

3. Final Answer:

   User: anonymous -> Option A

4. Quick Check:

   Empty variable triggers default value [OK]

Hint: Empty or unset triggers default with :- [OK]

Common Mistakes:

• Thinking empty string is treated as set
• Expecting empty output instead of default
• Confusing unset and empty behavior

4. Identify the error in this script snippet:

echo "Name: ${name-default}"

What is the problem?

medium

A. The dash should be a plus sign: ${name+default}.

B. The syntax is correct and will work as expected.

C. It should be ${name:-default} to use default if name is empty or unset.

D. The variable name is missing a $ sign.

Solution

1. Step 1: Understand the difference between - and :-

   Using ${name-default} uses default only if name is unset, but not if empty.

2. Step 2: Correct syntax for empty or unset

   To cover both unset and empty, ${name:-default} is needed.

3. Final Answer:

   It should be ${name:-default} to use default if name is empty or unset. -> Option C

4. Quick Check:

   Use :- to cover empty and unset [OK]

Hint: Use :- not - to cover empty and unset [OK]

Common Mistakes:

• Using - which ignores empty variables
• Confusing + and - operators
• Assuming syntax is always correct

5. You want to print the value of path if set and not empty, otherwise print "/usr/local/bin". But if path is set to an empty string, you want to treat it as unset. Which command correctly does this?

hard

A. echo ${path-/usr/local/bin}

B. echo ${path:-/usr/local/bin}

C. echo ${path:=/usr/local/bin}

D. echo ${path:+/usr/local/bin}

Solution

1. Step 1: Understand the requirement

   We want to print path if it is set and not empty; otherwise, print the default "/usr/local/bin".

2. Step 2: Choose correct parameter expansion

   ${path:-default} uses default if path is unset or empty. ${path-default} only uses default if unset, not empty. ${path:=default} assigns default to path. ${path:+default} prints default only if path is set.

3. Final Answer:

   echo ${path:-/usr/local/bin} -> Option B

4. Quick Check:

   :- covers unset and empty [OK]

Hint: Use :- to treat empty as unset and provide default [OK]

Common Mistakes:

• Using - which ignores empty strings
• Using := which changes variable value
• Using + which prints default only if set