Closures are reference types in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Swift code using closures?

Consider the following Swift code. What will be printed?

Swift

func makeIncrementer() -> () -> Int {
    var total = 0
    let incrementer: () -> Int = {
        total += 1
        return total
    }
    return incrementer
}

let increment = makeIncrementer()
print(increment())
print(increment())

A 2\n3

B 0\n1

C 1\n1

D 1\n2

💡 Hint

Remember that closures capture variables by reference, so the variable total keeps its value between calls.

The closure incrementer captures the variable total by reference. Each time increment() is called, total increases by 1. So the first call prints 1, the second prints 2.

🧠 Conceptual

intermediate

Why are closures considered reference types in Swift?

Choose the best explanation why closures in Swift behave as reference types.

A Because closures capture variables and share the same instance when assigned or passed around.

B Because closures are copied every time they are assigned to a new variable.

C Because closures cannot capture variables from their surrounding context.

D Because closures are stored on the stack and destroyed after use.

💡 Hint

Think about what happens when you assign a closure to multiple variables.

Closures capture variables from their surrounding context and share the same instance when assigned or passed around, so they behave like reference types.

🔧 Debug

advanced

What error occurs in this Swift closure code?

Examine the code below. What error will it produce when compiled?

Swift

func createClosure() -> () -> Int {
    var count = 0
    return {
        count += 1
        return count
    }
}

let closure1 = createClosure()
let closure2 = closure1
closure2 = createClosure()

A Type mismatch: cannot assign closure to closure2

B No error, code compiles and runs

C Cannot assign to value: 'closure2' is a 'let' constant

D Runtime error: variable 'count' not captured

💡 Hint

Look at how closure2 is declared and used.

closure2 is declared with let, so it cannot be assigned a new value after initialization. The line closure2 = createClosure() causes a compile-time error.

❓ Predict Output

advanced

What is the output when modifying a captured variable through two closures?

What will this Swift code print?

Swift

func makeCounter() -> () -> Int {
    var count = 0
    return {
        count += 1
        return count
    }
}

let counter1 = makeCounter()
let counter2 = counter1
print(counter1())
print(counter2())

A 1\n1

B 1\n2

C 2\n3

D 0\n1

💡 Hint

Remember closures are reference types and share captured variables.

Both counter1 and counter2 refer to the same closure instance, so the count variable is shared. The first call increments to 1, the second increments to 2.

🚀 Application

expert

How many unique closure instances are created here?

Given the code below, how many unique closure instances exist after execution?

Swift

func createClosure() -> () -> Int {
    var value = 0
    return {
        value += 1
        return value
    }
}

let c1 = createClosure()
let c2 = createClosure()
let c3 = c1
let c4 = c2
let c5 = c1

A 2

B 3

C 5

D 4

💡 Hint

Closures created by separate calls to createClosure() are unique instances.

Each call to createClosure() creates a new closure instance. c1 and c3 and c5 refer to the same instance, and c2 and c4 refer to another. So there are 2 unique closure instances.

Practice

1. What does it mean when we say closures are reference types in Swift?

easy

A. Multiple variables can point to the same closure instance, sharing changes.

B. Closures cannot capture variables from their surrounding context.

C. Each time you assign a closure, a new copy is created.

D. Closures are stored only as values, not references.

Solution

1. Step 1: Understand what reference types mean

   Reference types mean multiple variables can point to the same instance in memory.

2. Step 2: Apply this to closures

   Since closures are reference types, assigning them copies the reference, not the code itself.

3. Final Answer:

   Multiple variables can point to the same closure instance, sharing changes. -> Option A

4. Quick Check:

   Closures share instance references [OK]

Hint: Reference types share instances, not copies [OK]

Common Mistakes:

• Thinking closures always create new copies
• Confusing closures with value types like structs
• Ignoring that closures capture variables by reference

2. Which of the following is the correct way to assign a closure to a variable in Swift?

easy

A. func closure = { print("Hello") }

B. let closure = { print("Hello") }

C. closure := { print("Hello") }

D. var closure = function printHello() {}

Solution

1. Step 1: Recall Swift closure syntax

   Closures are assigned using curly braces and can be stored in variables with let or var.

2. Step 2: Check each option

   let closure = { print("Hello") } uses correct syntax: let closure = { print("Hello") }.

3. Final Answer:

   let closure = { print("Hello") } -> Option B

4. Quick Check:

   Correct closure assignment syntax [OK]

Hint: Closures use curly braces and assign with let or var [OK]

Common Mistakes:

• Using := which is not Swift syntax
• Trying to declare closures like functions
• Using func keyword incorrectly for closures

3. What will be printed by this Swift code?

var closureA = { print("Count: 0") }
var closureB = closureA
closureA = { print("Count: 1") }
closureB()

medium

A. Count: 0

B. Count: 1

C. No output, error occurs

D. Count: 0 and Count: 1

Solution

1. Step 1: Understand closure assignment

   closureB is assigned closureA's original closure printing "Count: 0".

2. Step 2: Reassign closureA

   closureA is changed to a new closure printing "Count: 1", but closureB still points to the old closure.

3. Final Answer:

   Count: 0 -> Option A

4. Quick Check:

   closureB keeps original closure [OK]

Hint: Reassigning closureA doesn't change closureB's reference [OK]

Common Mistakes:

• Assuming closureB changes when closureA is reassigned
• Confusing closure references with copies
• Expecting both closures to print the same after reassignment

4. Identify the error in this Swift code involving closures:

var closure1 = { print("Hello") }
var closure2 = closure1
closure1 = { print("World") }
closure2 = closure1

medium

A. Closure syntax is invalid

B. Cannot assign closure1 to closure2 after reassignment

C. No error, code runs fine

D. Variables closure1 and closure2 must be constants

Solution

1. Step 1: Check closure assignments

   Both closure1 and closure2 are variables and can be reassigned closures.

2. Step 2: Validate syntax and assignments

   All closure assignments use correct syntax and valid variable usage.

3. Final Answer:

   No error, code runs fine -> Option C

4. Quick Check:

   Valid closure reassignment [OK]

Hint: Variables can be reassigned closures freely [OK]

Common Mistakes:

• Thinking reassignment causes errors
• Confusing closure syntax with function syntax
• Assuming closures must be constants

5. Given this Swift code, what will be printed?

func makeCounter() -> () -> Int {
    var count = 0
    let increment: () -> Int = {
        count += 1
        return count
    }
    return increment
}

let counter1 = makeCounter()
let counter2 = counter1
print(counter1())
print(counter2())

hard

A. 1 and 1

B. 0 and 1

C. 2 and 2

D. 1 and 2

Solution

1. Step 1: Understand closure capturing

   The closure increments and returns the same captured variable 'count'.

2. Step 2: Analyze reference sharing

   counter1 and counter2 reference the same closure instance, so increments affect both.

3. Step 3: Trace calls

   First call counter1() returns 1, second call counter2() increments count to 2 and returns 2.

4. Final Answer:

   1 and 2 -> Option D

5. Quick Check:

   Shared closure state increments count [OK]

Hint: Closures share captured variables and state [OK]

Common Mistakes:

• Thinking counter2 is independent copy
• Ignoring closure captures and shared state
• Assuming count resets on each call