In-out parameters for mutation in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Swift code using in-out parameter?

Consider this Swift function that doubles a number using an in-out parameter. What will be printed?

Swift

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

var myNumber = 10
doubleValue(&myNumber)
print(myNumber)

A 10

B Compilation error

C 0

D 20

💡 Hint

Remember that in-out parameters allow the function to modify the original variable.

The function doubles the value of myNumber by modifying it directly through the in-out parameter. So the printed value is 20.

❓ Predict Output

intermediate

What error occurs when calling a function with in-out parameter incorrectly?

What error will this Swift code produce?

Swift

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

let constantValue = 5
increment(&constantValue)

A Cannot pass immutable value as inout argument

B Runtime error: nil value

C No error, prints 6

D Syntax error: missing &

💡 Hint

In-out parameters require a variable that can be changed.

You cannot pass a constant (let) as an in-out parameter because it cannot be mutated.

🔧 Debug

advanced

Why does this Swift code cause a compile error with in-out parameters?

Identify the cause of the compile error in this Swift code:

Swift

func swapValues(_ a: inout Int, _ b: inout Int) {
    let temp = a
    a = b
    b = temp
}

var x = 3
swapValues(&x, &x)

A Cannot pass the same variable twice as inout parameters

B Missing & before second argument

C Function swapValues requires return type

D No error, swaps values correctly

💡 Hint

Think about what happens if you try to modify the same variable twice at once.

Swift forbids passing the same variable as multiple in-out arguments simultaneously because it causes conflicting access to the variable.

❓ Predict Output

advanced

What is the output of this nested in-out parameter mutation?

What will be printed by this Swift code?

Swift

func addOne(_ number: inout Int) {
    number += 1
}

func doubleAndAddOne(_ number: inout Int) {
    number *= 2
    addOne(&number)
}

var value = 4
doubleAndAddOne(&value)
print(value)

A 10

B 8

C 9

D Compilation error

💡 Hint

Follow the changes step by step: first double, then add one.

The value starts at 4, doubles to 8, then addOne adds 1, resulting in 9.

🧠 Conceptual

expert

What is the main reason Swift uses in-out parameters instead of returning mutated values?

Why does Swift use in-out parameters for mutation instead of just returning a new value?

A Because Swift does not support returning values from functions

B To allow functions to modify variables directly without copying large data structures

C To prevent any changes to variables outside the function

D Because in-out parameters are required for all functions in Swift

💡 Hint

Think about performance and memory when working with big data.

In-out parameters allow functions to modify variables directly, which can be more efficient than returning new copies, especially for large data.