Command-line arguments ($1, $2, ...) in Bash Scripting - Practice Problems & Coding Challenges

Challenge - 5 Problems

💻 Command Output

intermediate

What is the output of this script with arguments?

Consider the following bash script saved as args.sh. What will be the output if you run ./args.sh apple banana cherry?

Bash Scripting

#!/bin/bash

echo "First argument: $1"
echo "Second argument: $2"
echo "All arguments: $@"

A

First argument: apple
Second argument: cherry
All arguments: apple banana cherry

B

First argument: banana
Second argument: cherry
All arguments: apple banana cherry

C

First argument: apple
Second argument: banana
All arguments: apple banana cherry

D

First argument: apple
Second argument: banana
All arguments: apple

💡 Hint

Remember that $1 is the first argument, $2 the second, and $@ represents all arguments.

In bash, $1 is the first argument passed to the script, $2 the second, and $@ expands to all arguments separated by spaces.

💻 Command Output

intermediate

What does this script print when no arguments are given?

Given this script args_empty.sh, what is the output when run as ./args_empty.sh with no arguments?

Bash Scripting

#!/bin/bash

echo "Argument 1 is: $1"
echo "Number of arguments: $#"

A

Argument 1 is: null
Number of arguments: 0

B

Argument 1 is: $1
Number of arguments: 1

C

Argument 1 is: 
Number of arguments: 1

D

Argument 1 is: 
Number of arguments: 0

💡 Hint

When no arguments are passed, $1 is empty and $# is zero.

If no arguments are passed, $1 is empty (prints nothing) and $# is 0 indicating zero arguments.

📝 Syntax

advanced

Which option correctly prints the third argument?

You want to print the third command-line argument in a bash script. Which of these lines will do it correctly?

A echo $3

B echo $[3]

C echo ${3}

D echo $@3

💡 Hint

Use the positional parameter syntax for arguments.

In bash, $3 prints the third argument. ${3} is invalid syntax. $[3] is arithmetic expansion and prints 3. $@3 is invalid.

🔧 Debug

advanced

Why does this script print empty for $2?

This script is run as ./script.sh one two. Why does it print an empty line for the second argument?

Bash Scripting

#!/bin/bash

echo "Arg1: $1"
echo "Arg2: $ 2"

A Because $2 is not set when two arguments are passed.

B Because there is a space between $ and 2, so $ is treated as variable and 2 as text.

C Because $ 2 is correct syntax but the second argument is empty.

D Because the script needs quotes around $2 to print it.

💡 Hint

Check the spacing between $ and the number.

In bash, $2 must be written without spaces. Writing $ 2 means $ alone (empty variable) followed by 2 as text, so it prints empty line.

🚀 Application

expert

How many arguments does this script count?

Given this script count_args.sh, what number will it print when run as ./count_args.sh a "b c" d?

Bash Scripting

#!/bin/bash

count=0
for arg in "$@"; do
  count=$((count + 1))
done

echo $count

A 3

B 4

C 1

D 5

💡 Hint

Remember that quoted arguments count as one argument.

The script loops over all arguments in "$@" which preserves quoted arguments as single items. The arguments are: a, b c, d → total 3.

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