Function arguments ($1, $2 inside function) in Bash Scripting - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print the first argument passed to the function.

Bash Scripting

my_func() {
  echo [1]
}
my_func Hello

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 $2 when only one argument is passed.

Inside a bash function, $1 refers to the first argument passed to that function.

2fill in blank

medium

Complete the code to print the second argument passed to the function.

Bash Scripting

print_second() {
  echo [1]
}
print_second apple banana

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

A$#

B$2

C$3

D$1

Attempts:

3 left

💡 Hint

Common Mistakes

Using $1 instead of $2 to get the second argument.

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

Inside the function, $2 refers to the second argument passed to it.

3fill in blank

hard

Fix the error in the function to correctly print the first and second arguments.

Bash Scripting

show_args() {
  echo "First: [1]"
  echo "Second: $2"
}
show_args one two

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

A$1

B$2

C$0

D$#

Attempts:

3 left

💡 Hint

Common Mistakes

Using $2 for the first argument.

Using $0 which is the script name, not function argument.

The first argument inside the function is accessed with $1, not $2 or $0.

4fill in blank

hard

Fill both blanks to print the first and second arguments inside the function.

Bash Scripting

args() {
  echo "Arg1: [1]"
  echo "Arg2: [2]"
}
args red blue

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

A$1

B$2

C$3

D$#

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping $1 and $2.

Using $3 or $# instead of $1 or $2.

Use $1 for the first argument and $2 for the second argument inside the function.

5fill in blank

hard

Fill all three blanks to create a function that prints all three arguments passed to it.

Bash Scripting

print_all() {
  echo "First: [1]"
  echo "Second: [2]"
  echo "Third: [3]"
}
print_all one two three

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

A$1

B$2

C$3

D$#

Attempts:

3 left

💡 Hint

Common Mistakes

Using $# which gives the count, not the argument value.

Mixing up the order of $1, $2, and $3.

Use $1, $2, and $3 to access the first, second, and third arguments inside the function.

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