How to convert an Int to a Character in Swift

Question

I\'ve struggled and failed for over ten minutes here and I give in. I need to convert an Int to a Character in Swift and cannot solve it.

Question<

Answer 1

For helpful context, taking vacawama's and Nate Cook's UnicodeScalar to use -

 let startingValue = Int(UnicodeScalar("A").value)
 for i in 0..<26 {
    let itemStr = String(UnicodeScalar(i + startingValue))

    items.append("Item " + itemStr)
}

Answer 2

How to convert an Int to a Character in Swift

For the sake of future visitors, I am providing a basic answer to the question title rather than the details of the question itself.

It is a two step process. Convert the Int to a UnicodeScalar and then convert the UnicodeScalar to a Character.

let myInteger: Int = 97

// convert Int to a valid UnicodeScalar
guard let myUnicodeScalar = UnicodeScalar(myInteger) else {
    return
}

// convert UnicodeScalar to Character
let myCharacter = Character(myUnicodeScalar)

// results
print(myCharacter) // a

(source)

Or alternatively...

if let myUnicodeScalar = UnicodeScalar(97) 
    let myCharacter = Character(myUnicodeScalar)
}

See also

• How to express Strings in Swift using Unicode hexadecimal values (UTF-16)
• Working with Unicode code points in Swift

Answer 3

New and updated

for charac in Unicode.Scalar("A").value...Unicode.Scalar("Z").value {
    print(Unicode.Scalar(charac)!, terminator:" ")}

prints:

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

thx for the help from @vacawama, I like this version, for Swift(5) especially because of:

for charac in Unicode.Scalar("a").value...Unicode.Scalar("z").value {
    print(Unicode.Scalar(charac)!, terminator:" ")}

prints:

a b c d e f g h i j k l m n o p q r s t u v w x y z

and not having to look up, even tho we should know our unicode? haha etc...

Answer 4

try this

for i in 0...25
{
    let string = String(format: "%c", i+65) as String
    NSLog("%@", string)
}

Answer 5

So far I've come up with this:

for i in 0 ..< 26 {
    print(Character(UnicodeScalar(Int(UnicodeScalar("A").value) + i)))
}

If you're just trying to generate "A" to "Z", you can avoid the math and just do:

for c in UnicodeScalar("A").value...UnicodeScalar("Z").value {
    print(String(UnicodeScalar(c)))
}

Answer 6

Simply convert the integer into String, then convert string into Character

let number = 5
let numChar = Character(String(number))