I imagine you’re right, that they’re NString indexes packaged up into a
frustrating return type. After sleeping on it, though, I imagined that even if
complex grapheme clusters WERE to make count( attrStr.string ) return a
different result than attrStr.length, it would probably never be due to
whitespace. So if I go back to Charles Strstka’s original suggestion, where you
pull off one character at a time, convert it to a 1-Character string, and then
test for whitespace membership, I should be able to count leading and trailing
whitespace characters and then do math based on attrStr.length to create the
range.
Here’s my current playground:
import Cocoa
extension Character {
func isMemberOfSet( set:NSCharacterSet )
-> Bool
{
// The for loop only executes once;
// its purpose is to convert Character to a type
// you can actually do something with
for char in String( self ).utf16 {
if set.characterIsMember( char ) {
return true
}
}
return false
}
}
var result:NSRange
let whitespace = NSCharacterSet.whitespaceAndNewlineCharacterSet()
let attrStr = NSAttributedString( string:" Fourscore and seven years ago...
\n\n \t\t" )
let str = attrStr.string
var headCount = 0
var tailCount = 0
var startIx = str.startIndex
var endIx = str.endIndex
while endIx > startIx && str[ endIx.predecessor() ].isMemberOfSet( whitespace )
{
++tailCount
endIx = endIx.predecessor()
}
if endIx > startIx {
while str[ startIx ].isMemberOfSet( whitespace ) {
++headCount
startIx = startIx.successor()
}
let length = attrStr.length - ( headCount + tailCount )
result = NSRange( location:headCount, length:length )
} else {
// String was empty or all whitespace
result = NSRange( location:0, length:0 )
}
let resultString = attrStr.attributedSubstringFromRange( result )
—
Charles
On April 2, 2015 at 11:16:52 PM, Quincey Morris
(quinceymor...@rivergatesoftware.com) wrote:
On Apr 2, 2015, at 19:28 , Charles Jenkins <cejw...@gmail.com> wrote:
I can indeed call attrStr.string.rangeOfCharacterFromSet(). But in typical
Swift string fashion, the return type is as unfriendly as possible:
Range<String.Index>? — as if the NSString were a Swift string.
I finally read the whole of what you said here, and I had to run to a
playground to check:
import Cocoa
var strA = "Hello?, String”
var strB = "Hello?, String" as NSString
var strC = "Hello\u{1f650}, String”
var strD = "Hello\u{1f650}, NSString" as NSString
var rangeA =
strA.rangeOfCharacterFromSet(NSCharacterSet.whitespaceCharacterSet()) // {Some
“7..<8”}
var rangeB =
strB.rangeOfCharacterFromSet(NSCharacterSet.whitespaceCharacterSet()) // (7,1)
var rangeC =
strC.rangeOfCharacterFromSet(NSCharacterSet.whitespaceCharacterSet()) // {Some
“8..<9”}
var rangeD =
strD.rangeOfCharacterFromSet(NSCharacterSet.whitespaceCharacterSet()) // (8,1)
So, yes, these are NSString indexes all the way, even if the result is packaged
as a Range<String.Index>.
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