Dictionary creation and access in Swift - Step-by-Step Execution

Concept Flow - Dictionary creation and access

Start

↓

Create Dictionary

↓

Add Key-Value Pairs

↓

Access Value by Key

↓

Check if Key Exists?

No→Return nil or default

Yes↓

Use Value

↓

End

This flow shows how to create a dictionary, add key-value pairs, then access values safely by checking if the key exists.

Execution Sample

Swift

var fruits = ["apple": "red", "banana": "yellow"]
let appleColor = fruits["apple"]
let orangeColor = fruits["orange"]

Create a dictionary of fruits and colors, then access colors by keys.

Execution Table

Step	Action	Dictionary State	Access Key	Value Retrieved
1	Create dictionary with apple and banana	{"apple": "red", "banana": "yellow"}	-	-
2	Access value for key "apple"	{"apple": "red", "banana": "yellow"}	apple	"red"
3	Access value for key "orange" (not present)	{"apple": "red", "banana": "yellow"}	orange	nil

💡 Finished accessing keys; orange key not found returns nil

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3
fruits	empty	{"apple": "red", "banana": "yellow"}	{"apple": "red", "banana": "yellow"}	{"apple": "red", "banana": "yellow"}
appleColor	nil	nil	Optional("red")	Optional("red")
orangeColor	nil	nil	nil	nil

Key Moments - 2 Insights

Why does accessing fruits["orange"] return nil instead of an error?

Because the dictionary does not have the key "orange", Swift returns nil safely instead of crashing, as shown in execution_table row 3.

Can we add new key-value pairs after dictionary creation?

Yes, since 'fruits' is declared with 'var', it is mutable and you can add or change pairs after creation.

Visual Quiz - 3 Questions

Look at the execution table, what value is retrieved when accessing the key "apple"?

A nil

B "yellow"

C "red"

D "orange"

💡 Hint

Check execution_table row 2 under 'Value Retrieved'

At which step does the dictionary get created with initial values?

A Step 1

B Step 2

C Step 3

D No step creates it

💡 Hint

Look at execution_table row 1 under 'Action' and 'Dictionary State'

If we add fruits["orange"] = "orange" after step 3, what will fruits["orange"] return?

A nil

B "orange"

C "red"

D Error

💡 Hint

Recall that 'fruits' is mutable (var) and adding a key-value pair updates the dictionary

Concept Snapshot

Dictionary creation: var dict = ["key": "value", ...]
Access value: dict["key"] returns optional value
If key missing, returns nil safely
Use var for mutable dictionary
Use let for constant dictionary

Full Transcript

This example shows how to create a dictionary in Swift with keys and values. We start by making a dictionary named 'fruits' with two pairs: apple-red and banana-yellow. Then we access the value for the key 'apple' which returns 'red'. Next, we try to access 'orange' which is not in the dictionary, so Swift returns nil safely without error. The dictionary is mutable because it is declared with var, so we can add or change pairs later. This teaches how to create and access dictionaries safely in Swift.

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