Why control flow directs execution in PowerShell - Challenge Your Understanding

Challenge - 5 Problems

💻 Command Output

intermediate

What is the output of this PowerShell script with if-else?

Look at this script and tell what it prints when $value is 5.

PowerShell

$value = 5
if ($value -gt 10) {
  Write-Output "Greater than 10"
} else {
  Write-Output "10 or less"
}

A 10 or less

B Greater than 10

C No output

D Error: Variable not defined

💡 Hint

Think about whether 5 is greater than 10 or not.

The if condition checks if $value is greater than 10. Since 5 is not, the else block runs, printing '10 or less'.

🧠 Conceptual

intermediate

Why does the script skip the else block here?

Given this script, why does it print "Inside if" and skip the else block?

PowerShell

$x = 20
if ($x -eq 20) {
  Write-Output "Inside if"
} else {
  Write-Output "Inside else"
}

A Because $x is not defined

B Because else always runs after if

C Because $x equals 20, so the if condition is true

D Because Write-Output only works inside if

💡 Hint

Check the condition inside the if statement.

The if condition checks if $x equals 20. Since it does, the if block runs and the else block is skipped.

📝 Syntax

advanced

Which option correctly uses a switch statement to print 'Fruit' for apple and banana?

Choose the correct PowerShell switch syntax that prints 'Fruit' when the input is 'apple' or 'banana'.

PowerShell

switch ($item) {
  'apple' { Write-Output 'Fruit' }
  'banana' { Write-Output 'Fruit' }
  default { Write-Output 'Unknown' }
}

A

switch ($item) {
  'apple': Write-Output 'Fruit'
  'banana': Write-Output 'Fruit'
  default: Write-Output 'Unknown'
}

B

switch ($item) {
  'apple' { Write-Output 'Fruit' }
  'banana' { Write-Output 'Fruit' }
  default { Write-Output 'Unknown' }
}

C

switch ($item) {
  'apple' -> Write-Output 'Fruit'
  'banana' -> Write-Output 'Fruit'
  default -> Write-Output 'Unknown'
}

D

switch $item {
  'apple' => Write-Output 'Fruit'
  'banana' => Write-Output 'Fruit'
  default => Write-Output 'Unknown'
}

💡 Hint

PowerShell switch uses braces {} for code blocks, not arrows or colons.

Option B uses the correct PowerShell switch syntax with braces and Write-Output inside each case block.

🔧 Debug

advanced

What error does this script produce and why?

This script tries to use a while loop but has a mistake. What error will it produce?

PowerShell

$count = 3
while $count -gt 0 {
  Write-Output $count
  $count--
}

A SyntaxError: Missing parentheses around while condition

B Runtime error: Variable $count not found

C SyntaxError: Missing parentheses around condition

D No error, prints 3 2 1

💡 Hint

Check how the while condition is written in PowerShell.

PowerShell requires the while condition to be inside parentheses. Missing parentheses cause a syntax error.

🚀 Application

expert

How many times does this nested loop print 'Looping'?

Count how many times 'Looping' is printed by this script.

PowerShell

$outer = 2
$inner = 3
for ($i = 0; $i -lt $outer; $i++) {
  for ($j = 0; $j -lt $inner; $j++) {
    Write-Output 'Looping'
  }
}

A 5 times

B 2 times

C 3 times

D 6 times

💡 Hint

Multiply the number of outer loop runs by inner loop runs.

The outer loop runs 2 times, inner loop runs 3 times each. Total prints = 2 * 3 = 6.