set -u for undefined variable errors in Bash Scripting - Practice Problems & Coding Challenges

Challenge - 5 Problems

💻 Command Output

intermediate

What is the output when using set -u with an undefined variable?

Consider this bash script snippet with set -u enabled. What will be the output or error?

Bash Scripting

set -u

echo "Value: $UNDEFINED_VAR"

A bash: UNDEFINED_VAR: unbound variable

B Value:

C Value: $UNDEFINED_VAR

D No output, script runs silently

💡 Hint

Think about what set -u does when a variable is not set.

The set -u option causes the shell to treat any attempt to use an undefined variable as an error. So, referencing $UNDEFINED_VAR will cause the script to stop and print an error message.

💻 Command Output

intermediate

How to safely use an undefined variable with set -u?

Given set -u is enabled, which command safely prints the value of MY_VAR without error if it is undefined?

Bash Scripting

set -u

# Which echo command below works safely?

A echo "$MY_VAR"

B echo "${MY_VAR+default}"

C echo "${MY_VAR?default}"

D echo "${MY_VAR:-default}"

💡 Hint

Look for the syntax that provides a default value if the variable is unset.

The ${VAR:-default} syntax returns default if VAR is unset or null, preventing errors with set -u.

🔧 Debug

advanced

Why does this script fail with set -u enabled?

Examine the script below. Why does it fail when set -u is set?

Bash Scripting

set -u

function greet() {
  echo "Hello, $1!"
}

greet $NAME

A Because $NAME is undefined and used without quotes

B Because $NAME is undefined and <code>set -u</code> causes an error on undefined variables

C Because $1 is undefined inside the function

D Because functions cannot access variables outside their scope

💡 Hint

Check which variable is undefined and how set -u treats it.

The variable $NAME is not defined before calling greet $NAME. With set -u, referencing an undefined variable causes an error.

🚀 Application

advanced

Modify this script to avoid errors with set -u

Given the script below, which modification prevents errors when set -u is enabled and CONFIG_PATH might be unset?

Bash Scripting

set -u

if [ -f "$CONFIG_PATH" ]; then
  echo "Config found"
else
  echo "Config missing"
fi

A if [ -f "${CONFIG_PATH?}" ]; then

B if [ -f "$CONFIG_PATH" ] 2>/dev/null; then

C if [ -f "${CONFIG_PATH:-}" ]; then

D if [ -f "$CONFIG_PATH" ] || true; then

💡 Hint

Use parameter expansion to provide a safe default for unset variables.

The ${VAR:-} syntax expands to an empty string if VAR is unset, preventing set -u errors.

🧠 Conceptual

expert

What is the main purpose of set -u in bash scripting?

Choose the best description of what set -u does in a bash script.

A It treats unset variables as an error and exits the script when they are used.

B It enables debugging mode to print each command before execution.

C It disables all variable expansions to improve performance.

D It automatically initializes all variables to empty strings.

💡 Hint

Think about how set -u helps catch mistakes with variables.

set -u makes the shell exit with an error if you try to use a variable that has not been set. This helps catch bugs early.