Default values for input in Bash Scripting - Practice Problems & Coding Challenges

Challenge - 5 Problems

💻 Command Output

intermediate

What is the output of this Bash script using default values?

Consider the following Bash script snippet. What will it print when run without any arguments?

Bash Scripting

name=${1:-"Guest"}
echo "Hello, $name!"

A Hello, Guest!

B Hello, $1!

C Hello, !

D Syntax error

💡 Hint

Look at how the variable 'name' is assigned using ${1:-"Guest"}.

The syntax ${var:-default} means use 'var' if set and not empty; otherwise, use 'default'. Since no argument is passed, $1 is empty, so 'Guest' is used.

💻 Command Output

intermediate

What does this script output when argument is provided?

What will this Bash script print if run with the argument "Alice"?

Bash Scripting

user=${1:-"Unknown"}
echo "User: $user"

A User: Alice

B User: $1

C User: Unknown

D User:

💡 Hint

Check how the default value is used only if the argument is missing.

Since the argument "Alice" is provided, $1 is set, so the variable 'user' takes the value 'Alice'.

📝 Syntax

advanced

Which option correctly assigns a default value to a variable if input is empty?

You want to assign the variable 'color' to the first argument or default to 'blue' if the argument is empty or unset. Which option is correct?

A color=$1 || blue

B color=${1-blue}

C color=${1:=blue}

D color=${1:-blue}

💡 Hint

Remember that :- uses default if variable is unset or empty.

Option D uses ${1:-blue} which means use $1 if set and not empty; otherwise, use 'blue'. Option D uses - which only substitutes if unset, not empty. Option D assigns 'blue' to $1 which is not intended. Option D is invalid syntax.

🔧 Debug

advanced

Why does this script print an empty string instead of the default?

Given this script: ```bash input=$1 value=${input-"default"} echo "$value" ``` When run without arguments, it prints an empty line. Why?

A Because echo is missing quotes

B Because the quotes around default cause a syntax error

C Because input is set but empty, so ${input-"default"} does not use default

D Because input is not exported

💡 Hint

Check what happens when assigning input=$1 with $1 unset, and note the - operator (no colon).

When run without arguments, $1 is unset. However, `input=$1` assigns an empty string to `input`, so `input` is set but null. The `${input-"default"}` (note: `-` without `:`) uses the value of `input` because it only substitutes if unset, not if empty. Thus, empty is printed.

🚀 Application

expert

How many items are in the array after this script runs?

Consider this Bash script: ```bash args=(${@:-"default"}) echo ${#args[@]} ``` What number will it print if run without any arguments?

A 0

B 1

C The number of positional parameters

D Syntax error

💡 Hint

Think about how ${@:-"default"} behaves when no arguments are passed; ${@} expands to null.

When no arguments are passed, ${@} expands to null (empty), so ${@:-"default"} expands to "default". Then `args=("default")`, so `${#args[@]}` is 1.