Dictionary methods and default values in Swift - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the updateValue(_:forKey:) method do in a Swift dictionary?

It updates the value for the given key if it exists, or adds a new key-value pair if the key is not present. It returns the old value if it was replaced, or nil if the key was new.

beginner

How can you safely access a value from a Swift dictionary with a default value if the key is missing?

Use the subscript with default value syntax: dict[key, default: defaultValue]. This returns the value for the key if it exists, or the default value otherwise.

intermediate

What is the difference between dict[key] and dict[key, default: defaultValue] in Swift?

dict[key] returns an optional value that is nil if the key is missing. dict[key, default: defaultValue] returns a non-optional value, using the default if the key is missing.

beginner

How does the removeValue(forKey:) method work in Swift dictionaries?

It removes the key-value pair for the given key if it exists and returns the removed value. If the key is not found, it returns nil.

beginner

What happens when you assign a value to a dictionary key using subscript syntax in Swift?

If the key exists, the value is updated. If the key does not exist, a new key-value pair is added to the dictionary.

What does dict.updateValue(10, forKey: "a") return if the key "a" was not in the dictionary?

A 10

B nil

C An error

D The old value for "a"

It returns nil because there was no old value for the key "a" before the update.

How do you provide a default value when accessing a dictionary key in Swift?

A dict.defaultValue(forKey: key)

B dict.get(key, defaultValue)

C dict[key] ?? defaultValue

D dict[key, default: defaultValue]

The correct syntax is dict[key, default: defaultValue]. Option C is valid Swift syntax but uses optional coalescing, not dictionary default subscript.

What type does dict[key] return in Swift?

A Optional value

B Always nil

C Non-optional value

D Array

dict[key] returns an optional because the key might not exist in the dictionary.

Which method removes a key-value pair from a Swift dictionary?

A remove(key:)

B delete(key:)

C removeValue(forKey:)

D pop(key:)

The correct method to remove a key-value pair is removeValue(forKey:).

What happens if you assign a value to a new key in a Swift dictionary using subscript syntax?

A The key-value pair is added

B The dictionary throws an error

C Nothing happens

D The dictionary clears all entries

Assigning a value to a new key adds that key-value pair to the dictionary.

Explain how to update a value in a Swift dictionary and how to handle the old value.

Think about the method that returns the previous value when you change a dictionary entry.

Describe how to access a dictionary value with a default fallback in Swift.

It's a special subscript syntax that avoids optionals.

Practice

1. What does the updateValue(_:forKey:) method do in a Swift dictionary?

easy

A. It updates the value for a given key and returns the old value if it existed.

B. It removes the key and its value from the dictionary.

C. It returns the value for a key or a default if the key is missing.

D. It creates a new dictionary with updated keys.

Solution

1. Step 1: Understand updateValue method purpose

   The updateValue(_:forKey:) method changes the value for a specified key in the dictionary.

2. Step 2: Check return behavior

   This method returns the old value if the key existed, otherwise nil.

3. Final Answer:

   It updates the value for a given key and returns the old value if it existed. -> Option A

4. Quick Check:

   updateValue returns old value or nil [OK]

Hint: Remember updateValue returns old value if key exists [OK]

Common Mistakes:

• Confusing updateValue with removeValue
• Thinking it returns the new value
• Assuming it creates a new dictionary

2. Which of the following is the correct syntax to get a value from a dictionary with a default in Swift?

easy

A. let value = dict[key, default: 0]

B. let value = dict.get(key, default: 0)

C. let value = dict[key] ?? 0

D. let value = dict.value(forKey: key, default: 0)

Solution

1. Step 1: Recall Swift dictionary default value syntax

   Swift dictionaries allow accessing values with a default using dict[key, default: defaultValue].

2. Step 2: Compare options

   let value = dict[key, default: 0] matches the correct syntax. let value = dict[key] ?? 0 uses nil-coalescing but is not the dictionary default syntax.

3. Final Answer:

   let value = dict[key, default: 0] -> Option A

4. Quick Check:

   Use dict[key, default:] for default values [OK]

Hint: Use square brackets with default: for safe dictionary access [OK]

Common Mistakes:

• Using get() method which doesn't exist
• Confusing nil-coalescing with dictionary default syntax
• Using value(forKey:) which is for NSObject

3. What is the output of this Swift code?

var scores = ["Alice": 10, "Bob": 8]
let oldValue = scores.updateValue(12, forKey: "Alice")
let missingValue = scores.updateValue(5, forKey: "Charlie")
print(oldValue, missingValue, scores)

medium

A. nil Optional(5) ["Alice": 12, "Bob": 8, "Charlie": 5]

B. 10 5 ["Alice": 12, "Bob": 8, "Charlie": 5]

C. Optional(10) nil ["Alice": 12, "Bob": 8, "Charlie": 5]

D. Optional(12) Optional(5) ["Alice": 10, "Bob": 8, "Charlie": 5]

Solution

1. Step 1: Understand updateValue return values

   Updating "Alice" from 10 to 12 returns old value Optional(10). Adding "Charlie" returns nil because key was missing.

2. Step 2: Check dictionary after updates

   Dictionary now has "Alice":12, "Bob":8, and "Charlie":5.

3. Final Answer:

   Optional(10) nil ["Alice": 12, "Bob": 8, "Charlie": 5] -> Option C

4. Quick Check:

   updateValue returns old value or nil [OK]

Hint: updateValue returns old value wrapped in Optional or nil if new key [OK]

Common Mistakes:

• Expecting updateValue to return new value
• Forgetting Optional wrapping
• Not updating dictionary correctly

4. Identify the error in this Swift code snippet:

var dict = ["a": 1, "b": 2]
let removed = dict.removeValue(forKey: "c")
print(removed!)

medium

A. Dictionary keys must be integers, not strings.

B. Force unwrapping removed causes runtime crash if key missing.

C. removeValue(forKey:) does not exist for dictionaries.

D. print statement syntax is incorrect.

Solution

1. Step 1: Understand removeValue return type

   removeValue(forKey:) returns an Optional value. If key is missing, it returns nil.

2. Step 2: Analyze force unwrapping

   Force unwrapping removed! when key "c" is missing causes a runtime crash.

3. Final Answer:

   Force unwrapping removed causes runtime crash if key missing. -> Option B

4. Quick Check:

   Force unwrap nil causes crash [OK]

Hint: Avoid force unwrapping optional from removeValue without check [OK]

Common Mistakes:

• Assuming removeValue always returns a value
• Ignoring optional unwrapping
• Thinking removeValue removes all keys

5. You have a dictionary var inventory = ["apple": 3, "banana": 5]. Write code to safely add 2 to the count of "orange", even if it doesn't exist yet, using dictionary default values. What is the correct way?

hard

A. inventory["orange"] = inventory["orange"]! + 2

B. inventory["orange"] = inventory["orange"] ?? 0 + 2

C. inventory.updateValue(inventory["orange"]! + 2, forKey: "orange")

D. inventory["orange"] = inventory["orange", default: 0] + 2

Solution

1. Step 1: Understand safe addition with default

   Using inventory["orange", default: 0] returns 0 if "orange" key is missing.

2. Step 2: Increment value safely

   Adding 2 to this default value and assigning back to inventory["orange"] updates or creates the key safely.

3. Final Answer:

   inventory["orange"] = inventory["orange", default: 0] + 2 -> Option D

4. Quick Check:

   dict[key] = dict[key, default: 0] + value for safe increments [OK]

Hint: Use dict[key] = dict[key, default: 0] + amount to add safely [OK]

Common Mistakes:

• Force unwrapping missing keys causing crash
• Using += on read-only default subscript
• Using updateValue without checking for nil