Lesson 8 - Strings in Swift - Working with single characters

In the previous exercise, Solved tasks for Swift lesson 7, we've practiced our knowledge from previous lessons.

If you noticed some similarities between arrays and strings, you were absolutely onto something. For the others, it may be a surprise that a String is essentially an array of Characters and we can work with it like so.

First, we'll check out how it works by simply printing the character at a given position. Unfortunately, we can't just specify the character index as a number, we have to retrieve it first via the index() method. This method returns a special String.Index type through which we can get to the character.

let s = "Hello ICT.social"
print(s)
let index = s.index(s.startIndex, offsetBy: 2)
print(s[index])

The output:

Hello ICT.social
l

We can see that we can access characters of a string through the brackets as it was with the array. It may be disappointing that characters at the given positions are read-only in Swift, so we can't write the following:

// This code won't work
var s = "Hello, ICT.social"
let index = s.index(s.startIndex, offsetBy: 2)
s[index] = "l"
println(s)

Of course, there is a way to do it, but we'll go over it later. For now, we'll be just reading characters.

Character frequency analysis

Let's write a simple program that analyzes a given sentence for us. We'll search for the number of vowels, consonants and non-alphanumeric characters (e.g. space or !).

We'll hard-code the input string in our code, so we won't have to enter it again every time. Once the program is complete, we'll replace the string with readLine(). We'll iterate over characters using a loop. I should start out by saying that we won't focus as much on program speed here, we'll choose practical and simple solutions.

First, let's define vowels and consonants. We don't have to count non-alphanumeric characters since it'll be the string length minus the number of vowels and consonants. Since we don't want to deal with the letter case, uppercase/lower­case, we'll convert the entire string to lowercase at the start. Let's set up variables for the individual counters, and also, because it's a more complex code, we'll add comments.

// the string that we want to analyze
var s = "A programmer gets stuck in the shower because the instructions on the shampoo were: Lather, Wash, and Repeat."
print(s)
s = s.lowercased()

// counters initialization
var vowelsCount = 0
var consonantsCount = 0

// definition of character groups
let vowels = "aeiouy"
let consonants = "bcdfghjklmnpqrstvwxz"

// main loop
for c in s {

}

First of all, we prepare the string and convert it to lowercase. Then, we reset the counters. For the definition of characters groups, we only need ordinary Strings. The main loop iterates over each character in the String s. In each iteration of the loop the variable c will contain the current character.

Now let's increment the counters. For simplicity's sake, I'll focus on the loop instead of rewriting the code repeatedly:

// main loop
for c in s {
    if vowels.contains(c) {
        vowelsCount += 1
    } else if consonants.contains(c) {
        consonantsCount += 1
    }
}

We already know the contains() method on a String. As a parameter, it can take both a substring or a character. Firstly, we try to find the character c from our sentence in the String vowels and possibly increase their counter. If it's not included in vowels, we look in consonants and possibly increase their counter.

Now all we're missing is the printing, displaying text, part at the end:

var s = "A programmer gets stuck in the shower because the instructions on the shampoo were: Lather, Wash, and Repeat."
println(s)
s = s.lowercased()

// counters initialization
var vowelCount = 0
var consonantCount = 0

// definition of character groups
let vowels = "aeiouy"
let consonants = "bcdfghjklmnpqrstvwxyz"

// main loop
for c in s {
        if vowels.contains(c) {
                vowelCount += 1
        } else if consonants.contains(c) {
                consonantCount += 1
        }
}

print("Vowels: \(vowelCount)")
print("Consonants: \(consonantCount)")
print("Other characters: \(s.length - (vowelCount + consonantCount))")

The output:

A programmer gets stuck in the shower because the instructions on the shampoo were: Lather, Wash, and Repeat.
Vowels: 33
Consonants: 55
Non-alphanumeric characters: 21

That's it, we're done!

ASCII value

Maybe you've already heard about the ASCII table. Especially, in the MS-DOS era when there was practically no other way to store text. Individual characters were stored as numbers of the byte datatype, so of a range from 0 to 255. The system provided the ASCII table which had 256 characters and each ASCII code (numerical code) was assigned to one character.

Perhaps you understand why this method is no longer as relevant. The table simply could not contain all the characters of all international alphabets, now we use Unicode (UTF-8) encoding where characters are represented in a different way. In Swift, we have the option to work with ASCII values of individual characters. The main advantage is that the characters are stored in the table next to each other, alphabetically. For example, at the position 97 we can find "a", at 98 "b" etc. It's the same with numbers, but unfortunately, the accent characters are messed up.

Now, let's convert a character into its ASCII value and vice versa create the character according to its ASCII value. The code is a little bit complicated; we'll describe it in a moment.

var c: Character // character
c = "a"
// conversion from text to ASCII value
let optionalASCIIvalue = c.unicodeScalars.filter{$0.isASCII}.first?.value
// we got an Optional but we already know what to do with it
if let ASCIIvalue = optionalASCIIvalue {
    print("The character \(c) was converted to its ASCII value of \(ASCIIvalue)")
}

// conversion from an ASCII value to text
c = Character(UnicodeScalar(98))
print("The ASCII value of 98 was converted to its textual value of \(c)")

Because especially the conversion of the character to its ASCII value looks quite frightening, let's describe briefly what this line of code actually does:

let optionalASCIIvalue = c.unicodeScalars.filter{$0.isASCII}.first?.value

In Swift, Characters and Strings are represented by Unicode Scalars, a 21-bit representation of one character. It can be, for example, "a", digits, special characters or emojis. We get this value or values from the unicodeScalars property of a Character or String.

filter allows us to select ASCII characters only, since Unicode Scalars also contain other special characters. We pass $0 into the filter, which (in this case) represents one Unicode Scalar. Using the isASCII property, we simply ask whether this character is part of ASCII table which is more limited than the Unicode Scalars representation. Then we just get the value of the first element found. That's an Optional since it may happen we won't find anything.

The Caesar cipher

Let's create a simple program to encrypt text. If you've ever heard of the Caesar cipher, then you already know exactly what we're going to program. The text encryption is based on shifting characters in the alphabet by a certain fixed number of characters. For example, if we shift the word "hello" by 1 character forwards, we'd get "ifmmp". The user will be allowed to select the number of character shifts.

Let's get right into it! We need variables for the original text, the encrypted message, and the shift. Then, we need a loop iterating over each character and we'll print the encrypted message at the end. We'll hard-code the message into our code again, so we won't have to enter it over and over during the testing phase. Once we finish the program, we'll replace the contents of the variable with the readLine() method. The cipher doesn't work with accent characters, spaces and punctuation marks. We'll just assume the user won't enter them. Ideally, we should remove accent characters before the encryption, as well as anything except letters.

// variable initialization
let s = "blackholesarewheregoddividedbyzero"
print("Original message: \(s)")
var message : String = ""
var shift = 1

// loop iterating over characters
for c in s {

}

// printing
print("Encrypted message: \(message)")

We'll now move into the loop. We'll convert the character in c to its ASCII value, its ordinal value, increase the value by the number of shifts and cast it back to a character. This character will be added to the final message:

// variable initialization
let s = "blackholesarewheregoddividedbyzero"
print("Original message: \(s)")
var message : String = ""
var shift : UInt32 = 1

// loop iterating over characters
for c in s {
    let optionalASCIIvalue = c.unicodeScalars.filter{$0.isASCII}.first?.value
    if let ASCIIvalue = optionalASCIIvalue {
        let newCharacter = Character(UnicodeScalar(ASCIIvalue + shift)!)
        message += [newCharacter]
    }
}

// printing
print("Encrypted message: \(message)")

The output:

Original message: blackholesarewheregoddividedbyzero
Encrypted message: cmbdlipmftbsfxifsfhpeejwjefeczafsp

You can see we opened Optional 'by force' once but we know what we're doing Why are there brackets around newCharacter when we add it to our message? Swift lets us to use the += operator to add another String or an array of characters to the original String, but not a single character. That's why we created a temporary array of a single character.

Let's try it out! The result looks pretty good. However, we can see that the characters after "z" overflow to ASCII values of other characters (e.g. "{"). Therefore, the characters are no longer just alphanumeric, but other nasty characters. Let's enclose our characters as a cyclical pattern, so the shifting could flow smoothly from "z" to "a" and so on. We'll get by with a simple condition that decreases the ASCII value by the length of the alphabet so we'd end back up at "a".

We'll create a new variable above the loop with the ASCII value of "z":

let zASCIIvalue = "z".unicodeScalars.filter{$0.isASCII}.first!.value

And modify the inside of the loop like this:

let optionalASCIIvalue = c.unicodeScalars.filter{$0.isASCII}.first?.value
if let ASCIIvalue = optionalASCIIvalue {
    var shiftedCharacter = ASCIIvalue + shift

    if shiftedCharacter > zASCIIvalue {
        shiftedCharacter -= 26
    }
    let newCharacter = Character(UnicodeScalar(shiftedCharacter)!)
    message += [newCharacter]
}

If shiftedCharacter exceeds the ASCII value of "z", we reduce it by 26 characters (the number of characters in the English alphabet). The -= operator does the same as we would do with shiftedCharacter = shiftedCharacter - 26. It's simple and our program is now working properly. Notice that we don't use direct character codes anywhere. There's there's the "z" character value in the condition which we retrieved before (once is enough, we don't have to do it in every iteration of the loop) even though we could write 122 there directly. We did it this way so that our program is fully independent from explicit ASCII values, so it'd be clearer how it works. Try to code the decryption program as practice for yourself.

In the following exercise, Solved tasks for Swift lesson 8, we're gonna practice our knowledge from previous lessons.