Why variables store and reuse data in Bash Scripting - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to assign the value 'hello' to the variable greeting.

Bash Scripting

greeting=[1]

Drag options to blanks, or click blank then click option'

A'hello'

Bhello

Cgreeting

D"hello"

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning the value without quotes causes errors or unexpected behavior.

Using the variable name as the value instead of the string.

In bash, to assign a string value to a variable, you enclose the string in single or double quotes. Here, 'hello' is correct.

2fill in blank

medium

Complete the code to print the value stored in the variable greeting.

Bash Scripting

echo [1]

Drag options to blanks, or click blank then click option'

A$greeting

B"greeting"

Cgreeting

D'greeting'

Attempts:

3 left

💡 Hint

Common Mistakes

Printing the variable name without $ prints the name, not the value.

Using quotes around the variable name prints it literally.

To print the value of a variable in bash, prefix the variable name with a dollar sign ($).

3fill in blank

hard

Fix the error in the code to correctly assign the value 5 to the variable count.

Bash Scripting

count=[1]

Drag options to blanks, or click blank then click option'

A"5"

B'5'

C5

D$5

Attempts:

3 left

💡 Hint

Common Mistakes

Using $ before a number causes bash to look for a positional parameter.

Putting numbers in quotes is allowed but unnecessary here.

In bash, numbers can be assigned without quotes. Using $5 tries to get the fifth argument, which is incorrect here.

4fill in blank

hard

Fill both blanks to create a variable named name with value 'Alice' and then print it.

Bash Scripting

[1]='Alice'
echo [2]

Drag options to blanks, or click blank then click option'

Aname

BName

C$name

D$Name

Attempts:

3 left

💡 Hint

Common Mistakes

Using $ when assigning a variable causes errors.

Mismatching variable name case between assignment and usage.

Variable names are case-sensitive. Assign to 'name' without $ and print with $name to get the value.

5fill in blank

hard

Fill all three blanks to assign 'blue' to color, then assign color to favorite, and finally print favorite.

Bash Scripting

color=[1]
favorite=[2]
echo [3]

Drag options to blanks, or click blank then click option'

A"blue"

B$color

C$favorite

Dblue

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning variables with $ on the left side causes errors.

Forgetting $ when referencing variable values.

Assign 'blue' with quotes to color, assign favorite to the value of color using $color, then print favorite with $favorite.