No implicit fallthrough in switch in Swift - Mini Project: Build & Apply

No Implicit Fallthrough in Switch

📖 Scenario: Imagine you are creating a simple program that tells the type of a fruit based on its name. You will use a switch statement to check the fruit name and print a message. Swift's switch statements do not allow implicit fallthrough, so each case must be handled explicitly.

🎯 Goal: Build a Swift program that uses a switch statement without implicit fallthrough to print the type of fruit based on a given fruit name.

📋 What You'll Learn

Create a variable called fruit with the value "Apple".

Create a variable called fruitType and set it to an empty string.

Use a switch statement on fruit with cases for "Apple", "Banana", and "Orange".

Assign fruitType a string describing the fruit type inside each case.

Print the fruitType variable.

💡 Why This Matters

🌍 Real World

Switch statements are used in many apps to handle different options or commands clearly and safely without accidental code running.

💼 Career

Understanding Swift's switch behavior is important for iOS app developers to write clear and bug-free decision-making code.

1

Create the fruit variable

Create a variable called fruit and set it to the string "Apple".

Swift

# Create a variable called fruit and set it to "Apple"
# Your code here

Need a hint?

Use var fruit = "Apple" to create the variable.

2

Create the fruitType variable

Create a variable called fruitType and set it to an empty string "".

Swift

var fruit = "Apple"
# Create a variable called fruitType and set it to an empty string
# Your code here

Need a hint?

Use var fruitType = "" to create the variable.

3

Use switch to assign fruitType

Use a switch statement on the variable fruit. Add cases for "Apple", "Banana", and "Orange". Inside each case, assign fruitType a string describing the fruit type: "Apple is a pome fruit", "Banana is a berry", and "Orange is a citrus fruit" respectively. Remember, Swift does not allow implicit fallthrough, so do not add fallthrough.

Swift

var fruit = "Apple"
var fruitType = ""
# Use switch on fruit to assign fruitType for "Apple", "Banana", and "Orange"
# Your code here

Need a hint?

Use switch fruit { case "Apple": ... } and assign fruitType inside each case. Do not use fallthrough.

4

Print the fruitType

Write a print statement to display the value of the variable fruitType.

Swift

var fruit = "Apple"
var fruitType = ""
switch fruit {
case "Apple":
    fruitType = "Apple is a pome fruit"
case "Banana":
    fruitType = "Banana is a berry"
case "Orange":
    fruitType = "Orange is a citrus fruit"
default:
    fruitType = "Unknown fruit type"
}
# Print the fruitType variable
# Your code here

Need a hint?

Use print(fruitType) to show the result.