Trying to obtain ASCII value of individual characters found in String using Swift

Question

I'm trying to figure out how to obtain the numeric value of characters that are contained in a String object.

e.g.

let checkString = "ball"
for i in checkString.characters {
    var val = Int(String(i))!
    print(val)
}

I would like my output to be:

ASCII value of 'b'
ASCII value of 'a'
ASCII value of 'l'
ASCII value of 'l'

However, I get instead the error:

fatal error: unexpectedly found nil while unwrapping an Optional value

In Java, the equivalent code is:

int val = str.charAt(i)

Can anyone see where it is I'm going wrong?

So you want `a` to be treated as 1 and `z` as 26. How do you want to handle the many thousands of other possible characters that might be in a string? — rmaddy, Oct 08 '17 at 23:42
BTW - read the documentation for `Int?(String)` to see why your code is crashing. — rmaddy, Oct 08 '17 at 23:43
@rmaddy Ideally, yes. However, is it possible for me to obtain the ascii value instead which is what the java code returns? — syedfa, Oct 08 '17 at 23:45
ASCII only covers a tiny portion of the possible Unicode values you can have in a string. — rmaddy, Oct 08 '17 at 23:48
@syedfa If you want the ASCII value then ask for that in your question instead of asking to get 1 for `a`, etc. — rmaddy, Oct 08 '17 at 23:51
@rmaddy Thanks very much for clarifying. I apologize for any confusion. — syedfa, Oct 08 '17 at 23:56
Since you really want the ASCII value we can close this as a duplicate. — rmaddy, Oct 08 '17 at 23:57

score 0 · Answer 1 · answered Oct 08 '17 at 23:48

Based on this post What's the simplest way to convert from a single character String to an ASCII value in Swift? you could try this:

extension String {
    var asciiArray: [UInt32] {
        return unicodeScalars.filter{$0.isASCII}.map{$0.value}
    }
}

extension Character {
    var asciiValue: UInt32? {
        return     String(self).unicodeScalars.filter{$0.isASCII}.first?.value
    }
}

"ball".lowercased().asciiArray.forEach { print($0 - 96) }

Notice this will only work for lowered case string since the calculation is ASCII based, therefore 97 equals "a", 98 "b" and so on. If you want to adapt it to upper case strings you would need to use this line instead:

"UPpER".uppercased().asciiArray.forEach { print($0 - 64) }

Are you sure? I just tested your case in my playground and get 1 for both of them (`A` and `a`) — Mauricio Chirino, Oct 08 '17 at 23:52

Trying to obtain ASCII value of individual characters found in String using Swift

1 Answers1