Command-line arguments ($1, $2, ...) in Bash Scripting - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print the first command-line argument.

Bash Scripting

echo "First argument is: $[1]"

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

A$1

B$2

C$0

D$#

Attempts:

3 left

💡 Hint

Common Mistakes

Using $0 instead of $1 to get the first argument.

Using $# which gives the number of arguments, not the argument itself.

The first command-line argument is accessed using $1 in bash scripts.

2fill in blank

medium

Complete the code to print the second command-line argument.

Bash Scripting

echo "Second argument is: $[1]"

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

A$0

B$2

C$1

D$#

Attempts:

3 left

💡 Hint

Common Mistakes

Using $1 instead of $2 for the second argument.

Using $0 which is the script name.

The second command-line argument is accessed using $2 in bash scripts.

3fill in blank

hard

Fix the error in the code to print all arguments count.

Bash Scripting

echo "Number of arguments: $[1]"

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

A$1

B$*

C$0

D$#

Attempts:

3 left

💡 Hint

Common Mistakes

Using $1 which is the first argument, not the count.

Using $0 which is the script name.

The special variable $# holds the number of command-line arguments.

4fill in blank

hard

Fill both blanks to print the first and third arguments.

Bash Scripting

echo "First: $[1], Third: $[2]"

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

A$1

B$2

C$3

D$#

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up $2 and $3 for third argument.

Using $# which is argument count, not argument value.

Use $1 for the first argument and $3 for the third argument.

5fill in blank

hard

Fill all three blanks to print the script name, first argument, and total arguments count.

Bash Scripting

echo "Script: $[1], Arg1: $[2], Count: $[3]"

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

A$1

B$#

C$0

D$2

Attempts:

3 left

💡 Hint

Common Mistakes

Using $1 for script name instead of $0.

Using $2 instead of $1 for first argument.

Using $1 or $2 instead of $# for count.

Use $0 for script name, $1 for first argument, and $# for total arguments count.

Practice

1. What does $1 represent in a bash script?

easy

A. The last command-line argument passed to the script

B. The total number of arguments passed to the script

C. The name of the script file

D. The first command-line argument passed to the script

Solution

1. Step 1: Understand command-line arguments in bash

   In bash scripts, $1 refers to the first argument given after the script name.

2. Step 2: Differentiate from other special variables

   $# is the count of arguments, $0 is the script name, and $@ is all arguments.

3. Final Answer:

   The first command-line argument passed to the script -> Option D

4. Quick Check:

   $1 = first argument [OK]

Hint: Remember $1 is always the first argument after script name [OK]

Common Mistakes:

• Confusing $1 with $# (argument count)
• Thinking $1 is the script name
• Mixing $1 with $0 or $@

2. Which of the following is the correct way to access the third argument in a bash script?

easy

A. $0

B. $2

C. $3

D. $#

Solution

1. Step 1: Recall argument numbering in bash

   Arguments start from $1 for the first, so the third argument is $3.

2. Step 2: Identify incorrect options

   $2 is second argument, $# is count, $0 is script name.

3. Final Answer:

   $3 -> Option C

4. Quick Check:

   Third argument = $3 [OK]

Hint: Count arguments starting at $1, so third is $3 [OK]

Common Mistakes:

• Using $2 instead of $3 for third argument
• Confusing $# with argument value
• Using $0 which is script name

3. What will be the output of this script if run as ./script.sh apple banana cherry?

echo "First: $1"
echo "Second: $2"
echo "Count: $#"

medium

A. First: apple
Second: banana
Count: 3

B. First: apple
Second: cherry
Count: 3

C. First: ./script.sh
Second: apple
Count: 3

D. First: banana
Second: cherry
Count: 2

Solution

1. Step 1: Identify argument values

   Arguments are apple ($1), banana ($2), cherry ($3). Script name is $0.

2. Step 2: Understand output commands

   echo prints $1 as apple, $2 as banana, and $# as 3 (number of arguments).

3. Final Answer:

   First: apple Second: banana Count: 3 -> Option A

4. Quick Check:

   Arguments match output [OK]

Hint: Remember $1, $2 are first and second args; $# is count [OK]

Common Mistakes:

• Mixing script name ($0) with arguments
• Confusing argument order
• Miscounting number of arguments

4. Given this script snippet:

echo "Argument 1: $1"
echo "Argument 2: $2"

What is the error if you run the script without any arguments?

medium

A. Script prints empty lines for $1 and $2

B. No output is printed

C. Script throws a syntax error

D. Script prints the script name instead of arguments

Solution

1. Step 1: Understand missing arguments behavior

   If no arguments are passed, $1 and $2 are empty strings, not errors.

2. Step 2: Check script output

   echo prints "Argument 1: " and "Argument 2: " with nothing after colon.

3. Final Answer:

   Script prints empty lines for $1 and $2 -> Option A

4. Quick Check:

   Missing args = empty $1, $2 [OK]

Hint: No args means $1, $2 are empty, no errors [OK]

Common Mistakes:

• Expecting syntax or runtime error
• Thinking script name prints as argument
• Assuming no output at all

5. You want a bash script to print all arguments passed to it, each on a new line, using only $1, $2, etc. Which approach works best?

hard

A. Use a loop with a counter and access arguments by $1, $2, ... until counter > $#

B. Use shift command in a loop to access each argument and print it

C. Use $0 to print all arguments one by one

D. Use echo $@ to print all arguments on one line

Solution

1. Step 1: Understand how to access all arguments one by one

   Using shift moves arguments left, so $1 becomes next argument each time.

2. Step 2: Compare options

   Use a loop with a counter and access arguments by $1, $2, ... until counter > $# is complex and error-prone; Use echo $@ to print all arguments on one line prints all on one line; Use $0 to print all arguments one by one prints script name; Use shift command in a loop to access each argument and print it correctly loops and prints each argument on new line.

3. Final Answer:

   Use shift command in a loop to access each argument and print it -> Option B

4. Quick Check:

   Shift loops through args one by one [OK]

Hint: Use shift in loop to print each argument separately [OK]

Common Mistakes:

• Using $0 instead of arguments
• Trying to manually count without shift
• Printing all args on one line instead of separate lines