Dictionary creation and access in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create an empty dictionary named ages.

Swift

var ages: [String: Int] = [1]

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

A{}

B[:]

CArray()

D()

Attempts:

3 left

💡 Hint

Common Mistakes

Using curly braces {} is invalid syntax for a dictionary.

Using parentheses () or Array() creates other types, not dictionaries.

In Swift, an empty dictionary is created using [:]. This means no key-value pairs yet.

2fill in blank

medium

Complete the code to create a dictionary capitals with keys and values.

Swift

let capitals = [1]

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

A["France", "Paris", "Japan", "Tokyo"]

B("France": "Paris", "Japan": "Tokyo")

C{"France": "Paris", "Japan": "Tokyo"}

D["France": "Paris", "Japan": "Tokyo"]

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets.

Using curly braces which is invalid syntax for dictionaries.

Listing keys and values as a flat array.

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

3fill in blank

hard

Fix the error in accessing the value for key "apple" in the dictionary fruitColors.

Swift

let color = fruitColors[1]"apple"

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

A(

B.get(

C[

D{

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses or curly braces instead of square brackets.

Trying to use a method like .get() which is not available in Swift dictionaries.

In Swift, dictionary values are accessed using square brackets with the key inside.

4fill in blank

hard

Complete the code to create a dictionary scores with keys as player names and values as their scores.

Swift

var scores: [String: Int] = [1]

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

A["Alice": 10, "Bob": 15]

B{"Alice": 10, "Bob": 15}

C("Alice", 10, "Bob", 15)

D["Alice", 10, "Bob", 15]

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses or curly braces instead of square brackets.

Listing keys and values as a flat list without colons.

Use square brackets with key-value pairs separated by colons to create a dictionary.

5fill in blank

hard

Complete the code to safely access the value for key "cat" in animalSounds and provide a default if missing.

Swift

let sound = animalSounds["cat"][1]"unknown"

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

A[

B??

C:

D]

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting to close the square brackets.

Using colon instead of nil-coalescing operator for default values.

Not providing a default value causing optional type issues.

Access dictionary value with square brackets, then use the nil-coalescing operator ?? to provide a default. Close the brackets properly.

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