In-out parameters for mutation in Swift - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is an in-out parameter in Swift?

An in-out parameter allows a function to modify a variable passed to it, so the changes persist outside the function.

beginner

How do you declare an in-out parameter in a Swift function?

Use the inout keyword before the parameter type in the function definition.

beginner

How do you pass a variable as an in-out argument when calling a Swift function?

Use the ampersand & before the variable name to pass it as an in-out argument.

intermediate

Why can't you pass a constant or a literal as an in-out argument?

Because in-out parameters require the ability to modify the variable, constants and literals cannot be changed.

beginner

What happens if you forget to use & when passing an in-out argument?

The Swift compiler will give an error because it expects an explicit indication that the variable can be modified.

Which keyword is used to declare an in-out parameter in Swift?

A inout

B mutate

C var

D ref

The inout keyword marks a parameter that can be modified inside the function.

How do you pass a variable as an in-out argument when calling a function?

A With parentheses around the variable

B With an asterisk (*) before the variable

C Just the variable name

D With an ampersand (&) before the variable

You must use & before the variable name to indicate it can be modified.

Can you pass a constant as an in-out argument?

A Yes, always

B Only if the function returns a value

C No, because it cannot be modified

D Yes, but only if marked with <code>let</code>

Constants cannot be passed as in-out arguments because they cannot be changed.

What is the main purpose of in-out parameters?

A To allow a function to modify the caller's variable

B To return multiple values from a function

C To make parameters optional

D To improve performance by copying variables

In-out parameters let functions change the original variable passed in.

What error occurs if you forget the & when passing an in-out argument?

A Syntax error

B Compiler error

C Runtime crash

D No error, it works fine

Swift compiler requires the & to clearly show the variable can be modified.

Explain how in-out parameters work in Swift and why they are useful.

Think about how a function can change a variable outside its own scope.

Describe the syntax for declaring and using an in-out parameter in a Swift function.

Focus on the keywords and symbols needed.

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