How do I check to see if a character is a space character in Swift?

Question

In a program I'm writing, I need to check to see if a character is a space (" "). Currently have this as the conditional but it's not working. Any ideas? Thanks in advance.

for(var k = indexOfCharBeingExamined; k < lineBeingExaminedChars.count; k++){

    let charBeingExamined = lineBeingExaminedChars[lineBeingExaminedChars.startIndex.advancedBy(k)];

//operations

    if(String(charBeingExamined) == " "){

    //more operations

    }
}

You need to add how charBeingExamined is being declared and the context of your code — Leo Dabus, Feb 07 '16 at 22:42

Dov · Answer 1 · 2020-08-19T21:53:48.477

4

The code below is how I solved this problem with a functional approach in Swift. I made an extension (two, actually), but you could easily take the guts of the function and use it elsewhere.

extension String {
    
    var isWhitespace: Bool {
        guard !isEmpty else { return true }
        
        let whitespaceChars = NSCharacterSet.whitespacesAndNewlines
        
        return self.unicodeScalars
            .filter { !whitespaceChars.contains($0) }
            .count == 0
    }
    
}

extension Optional where Wrapped == String {
    
    var isNullOrWhitespace: Bool {
        return self?.isWhitespace ?? true
    }
    
}

edited Aug 19 '20 at 21:53

answered Mar 15 '18 at 13:04

Dov

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2

It's perhaps better to use `contains` instead of `filter`, since it's cheaper. – pronebird Feb 16 '19 at 11:22
1

In Swift `5.2` explicit types no longer seem to be required. – Evan R Aug 09 '20 at 09:47

score 3 · Answer 2 · answered Feb 07 '16 at 23:01

The following code works for me. Note that it's easier to just iterate over the characters in a string using 'for' (second example below):

var s = "X yz"
for var i = 0; i < s.characters.count; i++ {
    let x = s[s.startIndex.advancedBy(i)]
    print(x)
    print(String(x) == " ")
}

for c in s.characters {
    print(c)
    print(String(c) == " ")
}

Eendje · Answer 3 · 2016-02-07T23:21:56.347

String:

let origin = "Some string with\u{00a0}whitespaces" // \u{00a0} is a no-break space

Oneliner:

let result = origin.characters.contains { " \u{00a0}".characters.contains($0) }

Another approach:

let spaces = NSCharacterSet.whitespaceCharacterSet()
let result = origin.utf16.contains { spaces.characterIsMember($0) }

Output:

print(result) // true

Not sure what you want to do with the spaces, because then it could be a bit simpler.

score 0 · Answer 4 · answered Aug 21 '19 at 10:24

just in your code change " " -> "\u{00A0}"

for(var k = indexOfCharBeingExamined; k < lineBeingExaminedChars.count; k++){
    let charBeingExamined = lineBeingExaminedChars[lineBeingExaminedChars.startIndex.advancedBy(k)];
    if(String(charBeingExamined) == "\u{00A0}"){
        //more operations
    }
}

score 0 · Answer 5 · answered Aug 11 '20 at 09:33

To test just for whitespace:

func hasWhitespace(_ input: String) -> Bool {
    let inputCharacterSet = CharacterSet(charactersIn: input)
    
    return !inputCharacterSet.intersection(CharacterSet.whitespaces).isEmpty
}

To test for both whitespace and an empty string:

func hasWhitespace(_ input: String) -> Bool {
    let inputCharacterSet = CharacterSet(charactersIn: input)
    
    return !inputCharacterSet.intersection(CharacterSet.whitespaces).isEmpty || inputCharacterSet.isEmpty
}

How do I check to see if a character is a space character in Swift?

5 Answers5