Character and String types in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a variable named greeting with the value "Hello".

Swift

var greeting: String = [1]

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A"hello"

B"Hello"

CHello

D'Hello'

Attempts:

3 left

💡 Hint

Common Mistakes

Using single quotes instead of double quotes for strings.

Not using quotes at all.

In Swift, strings are enclosed in double quotes. So, "Hello" is the correct way to assign the string value.

2fill in blank

medium

Complete the code to create a Character constant named letter with the value 'A'.

Swift

let letter: Character = [1]

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AA

B'A'

C"A"

D"a"

Attempts:

3 left

💡 Hint

Common Mistakes

Using single quotes for characters.

Not using quotes at all.

Characters in Swift are also enclosed in double quotes, just like strings, but assigned to a Character type.

3fill in blank

hard

Fix the error in the code to concatenate two strings greeting and name with a space in between.

Swift

let fullGreeting = greeting [1] " " [2] name

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A+

B&

C+=

D&&

Attempts:

3 left

💡 Hint

Common Mistakes

Using & or && which are not string concatenation operators in Swift.

Using += in the middle of an expression.

In Swift, the plus sign (+) is used to join (concatenate) strings.

4fill in blank

hard

Fill both blanks to create a dictionary that maps each word to its length if the length is greater than 3.

Swift

let wordLengths = Dictionary(uniqueKeysWithValues: words.filter { word in word.[1] > 3 }.map { word in (word, word.[2]) })

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Acount

Bword.length

Clen(word)

Dword.size

Attempts:

3 left

💡 Hint

Common Mistakes

Using len(word) which is not valid in Swift.

Using word.length or word.size which do not exist in Swift.

In Swift, to get the length of a string, use the count property. So word.count gives the length.

5fill in blank

hard

Fill all three blanks to create a dictionary that maps uppercase words to their lengths if length is greater than 4.

Swift

let result = Dictionary(uniqueKeysWithValues: words.filter { word in word.[3] > 4 }.map { word in (word.[1](), word.[2]) })

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

Auppercased

Bcount

Dlowercased

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercased() instead of uppercased().

Using incorrect length property.

Use word.uppercased() to convert to uppercase, and word.count to get length. The condition checks if length is greater than 4.