Function arguments ($1, $2 inside function) in Bash Scripting - Practice Problems & Coding Challenges

Challenge - 5 Problems

💻 Command Output

intermediate

Output of function using positional arguments

What is the output of this Bash script?

Bash Scripting

my_func() {
  echo "First: $1, Second: $2"
}
my_func apple banana

A First: , Second:

B First: my_func, Second: apple

C First: banana, Second: apple

D First: apple, Second: banana

💡 Hint

Remember that $1 and $2 inside a function refer to the function's arguments.

Inside the function, $1 is the first argument passed to it, which is 'apple', and $2 is the second argument, 'banana'.

💻 Command Output

intermediate

Effect of shifting arguments inside a function

What will this script output?

Bash Scripting

my_func() {
  shift
  echo "$1 $2"
}
my_func one two three

A three

B one two

C two three

D one three

💡 Hint

The shift command moves all arguments one position to the left.

After shift, $1 becomes 'two' and $2 becomes 'three'.

📝 Syntax

advanced

Identify the syntax error in function argument usage

Which option contains a syntax error when trying to print the first argument inside a function?

Bash Scripting

my_func() {
  echo $1
}
my_func test

A echo $[1]

B echo "$1"

C echo ${1}

D echo $1

💡 Hint

Check how positional parameters are referenced in Bash.

Option A uses $[1], which is arithmetic expansion syntax and invalid for positional parameters.

🔧 Debug

advanced

Why does this function print empty values?

Given this script, why does the function print empty lines?

Bash Scripting

my_func() {
  echo "$1 $2"
}
my_func

A Function syntax is incorrect

B No arguments were passed, so $1 and $2 are empty

C Variables $1 and $2 are not accessible inside functions

D The echo command is missing quotes

💡 Hint

Think about what happens if you call a function without arguments.

Since no arguments are passed, $1 and $2 inside the function are empty strings.

🚀 Application

expert

How to pass all script arguments to a function preserving spaces

You want to pass all arguments given to the script into a function, preserving spaces in each argument. Which option correctly does this?

Bash Scripting

my_func() {
  for arg in "$@"; do
    echo "$arg"
  done
}
# How to call my_func with all script arguments?

A my_func "$@"

B my_func "$*"

C my_func $@

D my_func $*

💡 Hint

Use quotes to preserve each argument as a separate word.

Using "$@" passes each argument as a separate quoted string, preserving spaces correctly.

Practice

1. In a bash function, what does $1 represent?

easy

A. The first argument passed to the function

B. The name of the function

C. The total number of arguments passed to the script

D. The exit status of the last command

Solution

1. Step 1: Understand function arguments in bash

   Inside a bash function, $1 refers to the first argument passed specifically to that function, not the script.

2. Step 2: Differentiate from script arguments

   The script's arguments are accessed outside functions, but inside a function, $1 resets to the function's first argument.

3. Final Answer:

   The first argument passed to the function -> Option A

4. Quick Check:

   $1 in function = first function argument [OK]

Hint: Inside functions, $1 is the first function argument, not script's [OK]

Common Mistakes:

• Confusing $1 inside function with script's $1
• Thinking $1 is function name
• Mixing $1 with number of arguments

2. Which of the following is the correct way to define a bash function that uses arguments $1 and $2?

easy

A. function myfunc { echo $1 $2

B. function myfunc($1, $2) { echo $1 $2 }

C. def myfunc($1, $2) { echo $1 $2 }

D. myfunc() { echo $1 $2; }

Solution

1. Step 1: Recall bash function syntax

   Bash functions are defined using name() { commands; } or function name { commands; } without parameters in parentheses.

2. Step 2: Identify correct syntax for arguments

   Arguments are accessed inside the function via $1, $2, not declared in parentheses.

3. Final Answer:

   myfunc() { echo $1 $2; } -> Option D

4. Quick Check:

   Function args accessed inside, no params in definition [OK]

Hint: Define functions with () and access args inside, no params in () [OK]

Common Mistakes:

• Trying to declare parameters in function definition
• Using other language syntax like def or parentheses with args
• Missing semicolon or braces

3. What is the output of this script?

greet() {
  echo "Hello, $1 and $2!"
}
greet Alice Bob

medium

A. Hello, $1 and $2!

B. Hello, Alice and Bob!

C. Hello, Alice and !

D. Hello, and Bob!

Solution

1. Step 1: Understand function call and arguments

   The function greet is called with two arguments: Alice and Bob. Inside the function, $1 is Alice and $2 is Bob.

2. Step 2: Substitute arguments in echo

   The echo command prints: "Hello, Alice and Bob!" exactly as expected.

3. Final Answer:

   Hello, Alice and Bob! -> Option B

4. Quick Check:

   Function args $1=Alice, $2=Bob replaced correctly [OK]

Hint: Arguments replace $1, $2 inside function call [OK]

Common Mistakes:

• Expecting literal $1 and $2 printed
• Mixing argument order
• Forgetting to pass both arguments

4. Consider this function:

sum() {
  echo $(($1 + $2))
}
sum 5

What error or output will this produce?

medium

A. bash: + : syntax error: operand expected (error token is "+ ")

B. 5

C. bash: syntax error: unexpected end of file

D. 10

Solution

1. Step 1: Analyze function call with missing argument

   The function sum expects two arguments, but only one (5) is passed. So, $2 is empty.

2. Step 2: Evaluate arithmetic expression

   The expression $(($1 + $2)) becomes (5 + ), which is invalid syntax causing an error about operand expected.

3. Final Answer:

   bash: + : syntax error: operand expected (error token is "+ ") -> Option A

4. Quick Check:

   Missing $2 causes arithmetic syntax error [OK]

Hint: Always pass all args used in arithmetic to avoid errors [OK]

Common Mistakes:

• Assuming missing args default to zero
• Ignoring error messages
• Passing arguments incorrectly

5. You want a bash function concat that joins two strings passed as arguments with a dash between them. Which function definition correctly does this?

hard

A. concat() { echo "$1+$2"; }

B. concat() { echo "$1 $2"; }

C. concat() { echo "$1-$2"; }

D. concat() { echo "$1$2"; }

Solution

1. Step 1: Understand the requirement

   The function should join two strings with a dash - between them.

2. Step 2: Check each option's output

   concat() { echo "$1-$2"; } prints $1-$2 which is exactly the two strings joined by a dash. Others use space, plus, or no separator.

3. Final Answer:

   concat() { echo "$1-$2"; } -> Option C

4. Quick Check:

   Dash between args = concat() { echo "$1-$2"; } [OK]

Hint: Use echo "$1-$2" to join with dash inside function [OK]

Common Mistakes:

• Using space or plus instead of dash
• Forgetting quotes causing word splitting
• Concatenating without separator