Accessing variables ($var and ${var}) in Bash Scripting - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does $var mean in bash scripting?

It means to get the value stored in the variable named var. For example, if var holds "hello", then $var will be replaced by "hello".

beginner

Why use ${var} instead of just $var?

Using ${var} helps bash know exactly where the variable name ends. This is useful when you want to add text right after the variable without spaces. For example, ${var}text clearly separates the variable from the word "text".

beginner

What happens if you write $vartext instead of ${var}text?

Bash will look for a variable named vartext, which might not exist. This can cause errors or unexpected results. Using ${var}text avoids this problem.

beginner

How do you print the value of a variable named name in bash?

You can use echo $name or echo ${name}. Both will print the value stored in name.

beginner

Can you use curly braces {} with variables in all cases?

Yes, you can always use ${var} safely. It is especially helpful when you want to add characters immediately after the variable name without confusion.

What does $var do in bash?

A Create a new variable named var

B Access the value stored in variable var

C Delete the variable var

D Print the word '$var' literally

$var accesses the value stored in the variable named var.

Why use ${var} instead of $var?

A To create a new variable

B To make the script run faster

C To clearly mark the variable name boundaries

D To comment out the variable

${var} helps bash know exactly where the variable name ends, especially when adding text right after it.

What happens if you write $vartext when you meant ${var}text?

A Bash looks for a variable named vartext

B Bash prints the value of var and then 'text'

C Bash throws a syntax error

D Bash ignores the text after var

Bash tries to find a variable named vartext, which may not exist, causing errors or wrong output.

How to print the value of variable name in bash?

A echo $name

B print name

C echo name

D print $name

echo $name prints the value stored in the variable name.

Is ${var} always safe to use?

A Only in functions

B No, it causes errors

C Only in loops

D Yes, it clearly defines variable boundaries

${var} is always safe and helps avoid confusion in variable names.

Explain how and why to use $var and ${var} in bash scripting.

Think about how bash reads variable names and what happens when you add text right after a variable.

Describe a common mistake when accessing variables without curly braces and how to fix it.

Focus on how bash interprets variable names when followed by text.

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