Get nth character of a string in Swift programming

Question

How can I get the nth character of a string? I tried bracket([]) accessor with no luck.

var string = "Hello, world!"

var firstChar = string[0] // Throws error

ERROR: 'subscript' is unavailable: cannot subscript String with an Int, see the documentation comment for discussion

Answer 1

My solution is in one line, supposing cadena is the string and 4 is the nth position that you want:

let character = cadena[advance(cadena.startIndex, 4)]

Simple... I suppose Swift will include more things about substrings in future versions.

Answer 2

I think that a fast answer for get the first character could be:

let firstCharacter = aString[aString.startIndex]

It's so much elegant and performance than:

let firstCharacter = Array(aString.characters).first

But.. if you want manipulate and do more operations with strings you could think create an extension..here is one extension with this approach, it's quite similar to that already posted here:

extension String {
var length : Int {
    return self.characters.count
}

subscript(integerIndex: Int) -> Character {
    let index = startIndex.advancedBy(integerIndex)
    return self[index]
}

subscript(integerRange: Range<Int>) -> String {
    let start = startIndex.advancedBy(integerRange.startIndex)
    let end = startIndex.advancedBy(integerRange.endIndex)
    let range = start..<end
    return self[range]
}

}

BUT IT'S A TERRIBLE IDEA!!

The extension below is horribly inefficient. Every time a string is accessed with an integer, an O(n) function to advance its starting index is run. Running a linear loop inside another linear loop means this for loop is accidentally O(n2) — as the length of the string increases, the time this loop takes increases quadratically.

Instead of doing that you could use the characters's string collection.

Answer 3

In order to feed the subject and show swift subscript possibilities, here's a little string "substring-toolbox" subscript based

These methods are safe and never go over string indexes

extension String {
    // string[i] -> one string char
    subscript(pos: Int) -> String { return String(Array(self)[min(self.length-1,max(0,pos))]) }

    // string[pos,len] -> substring from pos for len chars on the left
    subscript(pos: Int, len: Int) -> String { return self[pos, len, .pos_len, .left2right] }

    // string[pos, len, .right2left] -> substring from pos for len chars on the right
    subscript(pos: Int, len: Int, way: Way) -> String { return self[pos, len, .pos_len, way] }

    // string[range] -> substring form start pos on the left to end pos on the right
    subscript(range: Range<Int>) -> String { return self[range.startIndex, range.endIndex, .start_end, .left2right] }

    // string[range, .right2left] -> substring start pos on the right to end pos on the left
    subscript(range: Range<Int>, way: Way) -> String { return self[range.startIndex, range.endIndex, .start_end, way] }

    var length: Int { return countElements(self) }
    enum Mode { case pos_len, start_end }
    enum Way { case left2right, right2left }
    subscript(var val1: Int, var val2: Int, mode: Mode, way: Way) -> String {
        if mode == .start_end {
            if val1 > val2 { let val=val1 ; val1=val2 ; val2=val }
            val2 = val2-val1
        }
        if way == .left2right {
            val1 = min(self.length-1, max(0,val1))
            val2 = min(self.length-val1, max(1,val2))
        } else {
            let val1_ = val1
            val1 = min(self.length-1, max(0, self.length-val1_-val2 ))
            val2 = max(1, (self.length-1-val1_)-(val1-1) )
        }
        return self.bridgeToObjectiveC().substringWithRange(NSMakeRange(val1, val2))

        //-- Alternative code without bridge --
        //var range: Range<Int> = pos...(pos+len-1)
        //var start = advance(startIndex, range.startIndex)
        //var end = advance(startIndex, range.endIndex)
        //return self.substringWithRange(Range(start: start, end: end))
    }
}


println("0123456789"[3]) // return "3"

println("0123456789"[3,2]) // return "34"

println("0123456789"[3,2,.right2left]) // return "56"

println("0123456789"[5,10,.pos_len,.left2right]) // return "56789"

println("0123456789"[8,120,.pos_len,.right2left]) // return "01"

println("0123456789"[120,120,.pos_len,.left2right]) // return "9"

println("0123456789"[0...4]) // return "01234"

println("0123456789"[0..4]) // return "0123"

println("0123456789"[0...4,.right2left]) // return "56789"

println("0123456789"[4...0,.right2left]) // return "678" << because ??? range can wear endIndex at 0 ???

Answer 4

I just came up with this neat workaround

var firstChar = Array(string)[0]

Answer 5

Swift 4

Range and partial range subscripting using String's indices property

As variation of @LeoDabus nice answer, we may add an additional extension to DefaultBidirectionalIndices with the purpose of allowing us to fall back on the indices property of String when implementing the custom subscripts (by Int specialized ranges and partial ranges) for the latter.

extension DefaultBidirectionalIndices {
    subscript(at: Int) -> Elements.Index {
        return index(startIndex, offsetBy: at)
    }
}

// Moving the index(_:offsetBy:) to an extension yields slightly
// briefer implementations for these String extensions.
extension String {
    subscript(r: CountableClosedRange<Int>) -> SubSequence {
        return self[indices[r.lowerBound]...indices[r.upperBound]]
    }
    subscript(r: CountablePartialRangeFrom<Int>) -> SubSequence {
        return self[indices[r.lowerBound]...]
    }
    subscript(r: PartialRangeThrough<Int>) -> SubSequence {
        return self[...indices[r.upperBound]]
    }
    subscript(r: PartialRangeUpTo<Int>) -> SubSequence {
        return self[..<indices[r.upperBound]]
    }
}

let str = "foo bar baz bax"
print(str[4...6]) // "bar"
print(str[4...])  // "bar baz bax"
print(str[...6])  // "foo bar"
print(str[..<6])  // "foo ba"

Thanks @LeoDabus for the pointing me in the direction of using the indices property as an(other) alternative to String subscripting!

Swift 4.2.

In Swift 4.2, DefaultBidirectionalIndices has been deprecated in favour of DefaultIndices.

Answer 6

My very simple solution:

let myString = "Test string"
let index = 0
let firstCharacter = myString[String.Index(encodedOffset: index)]