Dictionary creation and access in Swift - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a dictionary in Swift?

A dictionary in Swift is a collection that stores key-value pairs. Each key is unique and is used to access its corresponding value.

beginner

How do you create an empty dictionary in Swift?

You create an empty dictionary by specifying the key and value types inside square brackets, like this:
var dict = [String: Int]()

beginner

How do you add or update a value in a Swift dictionary?

You assign a value to a key using subscript syntax:
dict["key"] = value. If the key exists, the value updates; if not, it adds a new pair.

intermediate

How do you safely access a value from a Swift dictionary?

Use optional binding to check if the value exists: <br><code>if let value = dict["key"] { /* use value */ }</code>. This avoids errors if the key is missing.

beginner

What happens if you access a key that does not exist in a Swift dictionary?

Accessing a non-existent key returns nil because dictionary values are optional when accessed by key.

How do you declare a dictionary with String keys and Int values in Swift?

A var dict = {String: Int}

B var dict = (String, Int)[]

C var dict = [String: Int]()

D var dict = [Int: String]()

Option C correctly declares an empty dictionary with String keys and Int values.

What does this code do?
dict["age"] = 30

A Creates a new dictionary

B Deletes the key "age"

C Accesses the value for key "age"

D Adds or updates the value 30 for the key "age"

Assigning a value to dict["age"] adds or updates the key-value pair.

What type is returned when you access a dictionary value by key in Swift?

A An optional of the value type

B The value type directly

C A Boolean

D A String

Accessing a dictionary by key returns an optional value because the key might not exist.

How can you safely use a value from a dictionary if the key might not exist?

A Use optional binding with if let

B Force unwrap the value

C Ignore the value

D Use a for loop

Optional binding lets you check if the value exists before using it safely.

Which of these is a correct way to create a dictionary with initial values?

A var dict = ("name" => "Alice", "age" => 25)

B var dict = ["name": "Alice", "age": 25]

C var dict = {"name": "Alice", "age": 25}

D var dict = ["name" -> "Alice", "age" -> 25]

Option B uses correct Swift syntax for dictionary literals.

Explain how to create a dictionary in Swift and add a new key-value pair.

Think about how you declare types and assign values.

Describe how to safely access a value from a Swift dictionary and handle the case when the key does not exist.

Remember that dictionary lookups return optionals.

Practice

1. What is a dictionary in Swift used for?

easy

A. To store data in key-value pairs for quick lookup

B. To store a list of values without keys

C. To perform mathematical calculations

D. To create user interface elements

Solution

1. Step 1: Understand dictionary purpose

   Dictionaries store data as pairs of keys and values, allowing fast access by key.

2. Step 2: Compare options

   Only To store data in key-value pairs for quick lookup correctly describes this key-value storage purpose.

3. Final Answer:

   To store data in key-value pairs for quick lookup -> Option A

4. Quick Check:

   Dictionaries = key-value storage [OK]

Hint: Remember: dictionaries pair keys with values for fast access [OK]

Common Mistakes:

• Confusing dictionaries with arrays (which have no keys)
• Thinking dictionaries perform calculations
• Assuming dictionaries create UI elements

2. Which of the following is the correct way to create a dictionary in Swift?

easy

A. let dict = ("apple", 1), ("banana", 2)

B. let dict = ["apple": 1, "banana": 2]

C. let dict = {"apple": 1, "banana": 2}

D. let dict = ["apple", 1, "banana", 2]

Solution

1. Step 1: Identify Swift dictionary syntax

   Swift dictionaries use square brackets with key-value pairs separated by colons.

2. Step 2: Check each option

   let dict = ["apple": 1, "banana": 2] uses correct syntax: ["key": value, ...]. Others use tuples, braces, or arrays incorrectly.

3. Final Answer:

   let dict = ["apple": 1, "banana": 2] -> Option B

4. Quick Check:

   Dictionary syntax = [key: value] [OK]

Hint: Use square brackets with colons for key-value pairs [OK]

Common Mistakes:

• Using parentheses instead of square brackets
• Using curly braces which define closures
• Listing keys and values without colons

3. What will be the output of this Swift code?

let fruits = ["apple": 3, "banana": 5]
print(fruits["banana"] ?? 0)

medium

A. 5

B. nil

C. banana

D. 0

Solution

1. Step 1: Access dictionary value by key

   fruits["banana"] looks up the value for key "banana", which is 5.

2. Step 2: Use nil-coalescing operator

   The ?? 0 means if the key is missing (nil), print 0; here key exists, so print 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   fruits["banana"] = 5 [OK]

Hint: Use ?? to provide default if key missing [OK]

Common Mistakes:

• Expecting nil instead of default value
• Printing the key instead of the value
• Confusing optional with direct value

4. Identify the error in this Swift code snippet:

var scores = ["Alice": 90, "Bob": 85]
scores["Charlie"] = 95
print(scores["Charlie"])

medium

A. Cannot add new key-value pairs to a dictionary

B. Dictionary keys must be integers

C. Syntax error in dictionary declaration

D. Missing unwrap of optional when printing

Solution

1. Step 1: Check dictionary modification

   Adding a new key-value pair to a var dictionary is allowed, so no error there.

2. Step 2: Understand optional access

   Accessing scores["Charlie"] returns an optional Int?; printing it directly shows Optional(95), not 95.

3. Step 3: Identify missing unwrap

   To print the value without Optional(...), unwrap it safely or use ?? to provide default.

4. Final Answer:

   Missing unwrap of optional when printing -> Option D

5. Quick Check:

   Dictionary access returns optional, unwrap needed [OK]

Hint: Remember dictionary access returns optional, unwrap before use [OK]

Common Mistakes:

• Thinking keys must be integers
• Assuming dictionary is immutable by default
• Ignoring optional nature of dictionary values

5. Given a dictionary let items = ["pen": 3, "notebook": 0, "eraser": 2], which Swift code correctly creates a new dictionary containing only items with quantity greater than zero?

hard

A. let filtered = items.compactMap { $0.value > 0 }

B. let filtered = items.filter { $0.key > 0 }

C. let filtered = items.filter { $0.value > 0 }

D. let filtered = items.map { $0.value > 0 }

Solution

1. Step 1: Understand filtering dictionary by value

   We want to keep pairs where value > 0, so filter with condition on $0.value.

2. Step 2: Analyze each option

   let filtered = items.filter { $0.value > 0 } uses filter with correct condition. let filtered = items.filter { $0.key > 0 } incorrectly compares key (a String) to 0. Options A and C use map/compactMap which transform values, not filter pairs.

3. Final Answer:

   let filtered = items.filter { $0.value > 0 } -> Option C

4. Quick Check:

   Filter dictionary by value > 0 = let filtered = items.filter { $0.value > 0 } [OK]

Hint: Use filter with $0.value condition to keep matching pairs [OK]

Common Mistakes:

• Filtering by key instead of value
• Using map instead of filter for selection
• Using compactMap which changes type, not filters