In-out parameters for mutation in Swift - Mini Project: Build & Apply

In-out Parameters for Mutation in Swift

📖 Scenario: You are working on a simple app that manages a user's bank account balance. Sometimes, you need to update the balance by adding or subtracting money directly.

🎯 Goal: Build a Swift program that uses an inout parameter to change the bank account balance inside a function.

📋 What You'll Learn

Create a variable called balance with the value 1000

Create a constant called depositAmount with the value 250

Write a function called deposit that takes an inout parameter named accountBalance of type Int and a parameter amount of type Int

Inside the function deposit, add amount to accountBalance

Call the function deposit passing balance as an inout argument and depositAmount as the amount

Print the updated balance

💡 Why This Matters

🌍 Real World

Banking apps and financial software often need to update account balances safely and clearly. Using <code>inout</code> parameters helps functions change values directly.

💼 Career

Understanding how to use <code>inout</code> parameters is important for Swift developers working on apps that manage data that changes, such as user profiles, game scores, or financial records.

1

Create the initial balance and deposit amount

Create a variable called balance and set it to 1000. Then create a constant called depositAmount and set it to 250.

Swift

// Create a variable called balance and set it to 1000
// Create a constant called depositAmount and set it to 250
// Your code here

Hint

Use var for balance because it will change. Use let for depositAmount because it stays the same.

2

Write the deposit function with an inout parameter

Write a function called deposit that takes an inout parameter named accountBalance of type Int and a parameter amount of type Int. Inside the function, add amount to accountBalance.

Swift

var balance = 1000
let depositAmount = 250

// Write the deposit function with inout parameter here
// Your code here

Hint

Remember to use inout before the parameter type to allow mutation inside the function.

3

Call the deposit function to update balance

Call the function deposit passing balance as an inout argument and depositAmount as the amount.

Swift

var balance = 1000
let depositAmount = 250

func deposit(accountBalance: inout Int, amount: Int) {
    accountBalance += amount
}

// Call the deposit function here
// Your code here

Hint

Use & before balance when passing it to the inout parameter.

4

Print the updated balance

Write a print statement to display the updated value of balance.

Swift

var balance = 1000
let depositAmount = 250

func deposit(accountBalance: inout Int, amount: Int) {
    accountBalance += amount
}

deposit(accountBalance: &balance, amount: depositAmount)

// Print the updated balance here
// Your code here

Hint

Use print(balance) to show the new balance.

Practice

1.

What does the inout keyword do in Swift functions?

easy

A. Allows the function to modify the original variable passed in.

B. Prevents the function from changing the variable.

C. Creates a copy of the variable that cannot be changed.

D. Makes the variable constant inside the function.

Solution

1. Step 1: Understand the role of inout

   The inout keyword allows a function to receive a variable and modify it directly.

2. Step 2: Effect on the original variable

   Changes made inside the function affect the original variable outside the function.

3. Final Answer:

   Allows the function to modify the original variable passed in. -> Option A

4. Quick Check:

   inout means modify original variable [OK]

Hint: Remember: inout means function can change your variable [OK]

Common Mistakes:

• Thinking inout creates a copy instead of modifying original
• Forgetting to use & when calling the function
• Assuming variables are constant inside the function

2.

Which of the following is the correct way to declare a function with an inout parameter in Swift?

func updateValue(____ value: inout Int) { value += 10 }

easy

A. func updateValue(inout value: Int)

B. func updateValue(value: inout Int)

C. func updateValue(value inout Int)

D. func updateValue(inout value in Int)

Solution

1. Step 1: Recall Swift syntax for inout parameters

   The correct syntax places inout before the type, after the parameter name: paramName: inout Type.

2. Step 2: Match the options

   func updateValue(value: inout Int) matches this syntax exactly: value: inout Int.

3. Final Answer:

   func updateValue(value: inout Int) -> Option B

4. Quick Check:

   Parameter syntax = name: inout Type [OK]

Hint: Remember: inout goes after parameter name, before type [OK]

Common Mistakes:

• Placing inout before parameter name
• Swapping parameter name and type order
• Using invalid syntax like 'inout value in Int'

3.

What will be the output of this Swift code?

func doubleValue(_ number: inout Int) {
    number *= 2
}

var myNum = 5
doubleValue(&myNum)
print(myNum)

medium

A. 5

B. Error: missing & in function call

C. 10

D. Error: cannot modify constant

Solution

1. Step 1: Understand the function behavior

   The function doubles the value of the passed variable by multiplying it by 2.

2. Step 2: Trace the code execution

   Variable myNum starts at 5, is passed with & to allow mutation, so after the function call it becomes 10.

3. Final Answer:

   10 -> Option C

4. Quick Check:

   In-out parameter doubles 5 to 10 [OK]

Hint: Use & when calling to let function change variable [OK]

Common Mistakes:

• Forgetting & when calling function
• Expecting original value unchanged
• Confusing parameter passing with copy

4.

Identify the error in this Swift code snippet:

func increment(_ value: inout Int) {
    value += 1
}

let number = 10
increment(&number)

medium

A. Cannot pass a constant as an inout parameter.

B. Missing & when calling the function.

C. Function parameter should not be inout.

D. Function must return the new value.

Solution

1. Step 1: Check variable mutability

   The variable number is declared with let, making it a constant.

2. Step 2: Understand inout parameter requirements

   Inout parameters require a variable that can be modified, so constants cannot be passed.

3. Final Answer:

   Cannot pass a constant as an inout parameter. -> Option A

4. Quick Check:

   Inout needs variable, not constant [OK]

Hint: Only variables (var), not constants (let), can be inout [OK]

Common Mistakes:

• Trying to pass constants as inout parameters
• Forgetting & when calling function
• Thinking function must return value when using inout

5.

You want to write a function swapValues that swaps two integers using inout parameters. Which of the following implementations correctly swaps the values?

func swapValues(_ a: inout Int, _ b: inout Int) {
    ____
}

hard

A. let temp = a
temp = b
b = temp

B. a = b
b = a

C. a = b
b = a - b
a = a - b

D. var temp = a
a = b
b = temp

Solution

1. Step 1: Understand swapping logic

   To swap two values, you need a temporary variable to hold one value while assigning the other.

2. Step 2: Check each option

   var temp = a a = b b = temp correctly declares a temporary variable with var and swaps values safely. let temp = a temp = b b = temp uses let which is immutable and causes error. a = b b = a overwrites a before assigning b. a = b b = a - b a = a - b uses arithmetic swap but is more complex and not recommended here.

3. Final Answer:

   var temp = a a = b b = temp -> Option D

4. Quick Check:

   Use temp variable with var to swap values [OK]

Hint: Use a temporary variable declared with var to swap values [OK]

Common Mistakes:

• Using let for temporary variable causing immutability error
• Overwriting one variable before saving its value
• Trying to swap without a temporary variable incorrectly