In-out parameters for mutation in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare an in-out parameter in the function.

Swift

func increment(number: [1] Int) {
    number += 1
}

Drag options to blanks, or click blank then click option'

Alet

Bvar

Cinout

Dmutating

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'var' instead of 'inout' to declare a mutable parameter.

Forgetting to use 'inout' when intending to modify the parameter.

In Swift, inout is used to allow a function to modify a parameter's value.

2fill in blank

medium

Complete the code to call the function with an in-out parameter correctly.

Swift

var value = 5
increment([1]value)

Drag options to blanks, or click blank then click option'

A#

B*

C$

D&

Attempts:

3 left

💡 Hint

Common Mistakes

Omitting the '&' symbol when passing the variable.

Using incorrect symbols like '*', '$', or '#'.

When passing an in-out parameter, you must prefix the variable with & to indicate it can be modified.

3fill in blank

hard

Fix the error in the function call by completing the blank.

Swift

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

var num = 10
doubleValue(number: [1]num)

Drag options to blanks, or click blank then click option'

A&

B#

C$

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Calling the function without the '&' symbol.

Using other symbols like '*', '$', or '#'.

The & symbol is required before the variable name when passing it as an in-out parameter.

4fill in blank

hard

Fill both blanks to create a function that swaps two integers using in-out parameters.

Swift

func swapInts(a: [1] Int, b: [2] Int) {
    let temp = a
    a = b
    b = temp
}

Drag options to blanks, or click blank then click option'

Ainout

Bvar

Clet

Dmutating

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'var' or 'let' instead of 'inout'.

Declaring only one parameter as 'inout'.

Both parameters must be declared as inout to allow swapping their values.

5fill in blank

hard

Fill all three blanks to create a function that increments a number by a given amount using in-out parameters.

Swift

func incrementBy(amount: [1] Int, number: [2] Int) {
    number [3]= amount
}

Drag options to blanks, or click blank then click option'

Alet

Binout

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Declaring amount as inout instead of let.

Using '-=' instead of '+='.

Not using inout for number.

The amount is a constant input (let), number is an inout parameter to be modified, and += adds the amount to number.

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