Shifting arguments (shift) in Bash Scripting - Practice Problems & Coding Challenges

Challenge - 5 Problems

💻 Command Output

intermediate

What is the output after shifting arguments?

Consider the following bash script snippet. What will be printed when the script is run with arguments: one two three?

Bash Scripting

echo "$1"
shift
echo "$1"

A one\ntwo

B two\nthree

C one\nthree

D three\nthree

💡 Hint

Remember that shift moves all arguments to the left by one position.

Initially, $1 is 'one'. After shift, $1 becomes the original $2, which is 'two'. So the outputs are 'one' then 'two'.

💻 Command Output

intermediate

How many arguments remain after shifting twice?

Given a script run with 4 arguments, what is the value of $# after executing shift 2?

Bash Scripting

shift 2
echo "$#"

A 3

B 4

C 0

D 2

💡 Hint

Each shift removes one argument from the front.

Starting with 4 arguments, shifting by 2 removes the first two, leaving 2 arguments.

📝 Syntax

advanced

Identify the syntax error in this shift usage

Which option contains a syntax error when trying to shift arguments by 2?

Bash Scripting

shift 2

A shift 2

B shift 0

C shift -2

D shift 1

💡 Hint

Shift count must be a non-negative integer.

Using a negative number with shift is invalid syntax and causes an error.

🚀 Application

advanced

What is the output of this script using shift in a loop?

Given the script below, what will it print when run with arguments a b c?

Bash Scripting

while [ "$#" -gt 0 ]; do
  echo "$1"
  shift
 done

A a\nb\nc

B a\na\na

C c\nb\na

D b\nc

💡 Hint

The loop prints $1 then shifts until no arguments remain.

The loop prints each argument in order, shifting after each print, so outputs 'a', then 'b', then 'c'.

🧠 Conceptual

expert

What happens if you shift more than the number of arguments?

If a script is run with 2 arguments and executes shift 3, what is the value of $# afterwards?

A 2

B 0

C -1

D 3

💡 Hint

Shifting more than available arguments removes all arguments without error.

Shifting more than the number of arguments sets $# to zero without error.