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.