On Apr 1, 2015, at 21:17 , Charles Jenkins <cejw...@gmail.com> wrote:
>
> for ch in String(char).utf16 {
> if !set.characterIsMember(ch) { found = false }
> }
Except that this code can’t possibly be right, in general.
1. A ‘unichar’ is a UTF-16 code value, but it’s not a Unicode code point. Some
UTF-16 code values have no meaning as “characters” by themselves. I think you
could mitigate this problem by using ‘longCharacterIsMember’, which takes a
UTF-32 code value instead (and enumerating the string as UTF-32 instead of
UTF-16).
2. A Swift ‘Character’ isn’t a Unicode code point, but rather a grapheme. That
is, it might be a sequence of code points (and I mean code points, not code
values). It might be such a sequence either because there’s no way of
representing the grapheme by a single code point, or because it’s a composed
character made up of a base code points and some combining characters.
In this case, you can’t validly test the individual code points for membership
of the character set.
I’m not sure, but I suspect the underlying obstacle is that NSCharacterSet is
at best a set of code points, and you cannot test a grapheme for membership of
a set of code points.
In your particular application, if it’s true that all** Unicode whitespace
characters are represented as a single code point (via a single UTF-32 code
value), or a single UTF-16 code value, then you can get away with one of the
above solutions. Otherwise you’re going to need a more complex solution, that
doesn’t involve NSCharacterSet at all.
** Or at least the ones you happen to care about, but ignoring the others may
be a perilous proceeding.
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