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.