Local variables (local keyword) in Bash Scripting - Practice Problems & Coding Challenges

Challenge - 5 Problems

💻 Command Output

intermediate

What is the output of this Bash script using local variables?

Consider the following Bash script. What will it print when run?

Bash Scripting

my_func() {
  local var="inside"
  echo "$var"
}
var="outside"
my_func
echo "$var"

A inside\noutside

B inside\ninside

C outside\noutside

D outside\ninside

💡 Hint

Think about how the local keyword limits the variable scope inside the function.

The local keyword creates a variable that exists only inside the function. So inside the function, var is 'inside'. Outside, var remains 'outside'.

💻 Command Output

intermediate

What error or output does this script produce?

What happens when you run this Bash script?

Bash Scripting

func() {
  local x=5
  ((x++))
  echo $x
}
func
echo $x

A 6\n\n

B 6\n

C 6\n5

D Syntax error

💡 Hint

Remember local variables do not affect global variables.

Inside func, x is local and incremented to 6 and printed. Outside, x is undefined, so echo prints an empty line.

📝 Syntax

advanced

Which option correctly declares a local variable with initial value in Bash?

Choose the correct syntax to declare a local variable named 'count' with value 10 inside a function.

A local count 10

B local count=10

C local count := 10

D local count == 10

💡 Hint

Bash uses = to assign values, no spaces around =.

In Bash, local variables are declared with 'local var=value' syntax. Other options are invalid syntax.

🔧 Debug

advanced

Why does this script print 'local' twice instead of 'local' and 'global'?

Examine the script below and find why the output is not as expected.

Bash Scripting

var="global"
func() {
  var="local"
  echo "$var"
}
func
echo "$var"

A Because the function is not called properly.

B Because 'var' is always global in Bash, local keyword is ignored.

C Because 'var' inside func is not declared local, it overwrites the global variable.

D Because echo cannot print local variables.

💡 Hint

Think about variable scope and the effect of missing 'local'.

Without 'local', the assignment inside func changes the global var. So both echoes print 'local'.

🚀 Application

expert

How many variables are accessible outside after running this script?

Given the script below, how many variables named 'a', 'b', and 'c' exist and are accessible outside the function?

Bash Scripting

a=1
func() {
  local a=2
  b=3
  local c=4
}
func
echo "$a $b $c"

A No variables accessible globally

B 3 variables: 'a', 'b', and 'c' all accessible globally

C 1 variable: only 'a' accessible globally

D 2 variables: 'a' and 'b' accessible; 'c' is local and not accessible

💡 Hint

Local variables exist only inside functions; others affect global scope.

'a' outside is global and unchanged. 'b' is assigned without local, so global. 'c' is local and not accessible outside.