CODEWITHSUNDEEP
swiftintcharacterswift-playgroundswift2
Aaron
Aaron

Reputation: 1342

Convert Character to Int in Swift 2.0

I just want to convert a character into an Int.

This should be simple. But I haven't found the previous answers helpful. There is always some error. Perhaps it is because I'm trying it in Swift 2.0. 

for i in (unsolved.characters) {
  fileLines += String(i).toInt()
  print(i)
}

Upvotes: 30

Views: 35302

Answers (5)

parametr
parametr

Reputation: 1102

In the latest Swift versions (at least in Swift 5) you don't need to do additional conversion to String.

Character has property wholeNumberValue which tries to convert a character to Int and returns nil if the character does not represent and integer.

let char: Character = "5"
if let intValue = char.wholeNumberValue {
    print("Value is \(intValue)")
} else {
    print("Not an integer")
}

Upvotes: 3

Declan McKenna
Declan McKenna

Reputation: 4870

Swift 4 solutions

To convert a Character to a String:

let digit = Int(String("1"))! 
//Don't force unwrap this if you're not 100% sure your character is a digit

Solution to OPs problem:

let fileLines = unsolved.compactMap{ Int(String($0) }
// fileLines = [1, 2, 3]

Upvotes: 3

MiladiuM
MiladiuM

Reputation: 1449

This is a bit of improvisational trick I came up with. Since it can be tricky to convert Character to Int, but you can easily convert String to Int do this :

fileLines = Int(String(i))!

of course this is not very well optimized but for cases like yours it can do the trick.

Upvotes: 9

Aaron Brager
Aaron Brager

Reputation: 66242

In Swift 2.0, toInt(), etc., have been replaced with initializers. (In this case, Int(someString).)

Because not all strings can be converted to ints, this initializer is failable, which means it returns an optional int (Int?) instead of just an Int. The best thing to do is unwrap this optional using if let.

I'm not sure exactly what you're going for, but this code works in Swift 2, and accomplishes what I think you're trying to do:

let unsolved = "123abc"

var fileLines = [Int]()

for i in unsolved.characters {
    let someString = String(i)
    if let someInt = Int(someString) {
        fileLines += [someInt]
    }
    print(i)
}

Or, for a Swiftier solution:

let unsolved = "123abc"

let fileLines = unsolved.characters.filter({ Int(String($0)) != nil }).map({ Int(String($0))! })

// fileLines = [1, 2, 3]

You can shorten this more with flatMap:

let fileLines = unsolved.characters.flatMap { Int(String($0)) }

flatMap returns "an Array containing the non-nil results of mapping transform over self"… so when Int(String($0)) is nil, the result is discarded.

Upvotes: 47

Tuan Anh Vu
Tuan Anh Vu

Reputation: 780

Actually. A simpler way is to convert String to Int in Swift 2.o is:

let chars = Int(chars)

Not sure if this is what you're trying for...But you can easily apply this to your loop, of course.

Upvotes: 3

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