Swift为什么将这个字素簇算作两个字符而不是一个？ [英] Why is Swift counting this Grapheme Cluster as two characters instead of one?
问题描述
通常，Swift关于将字素簇作为单个字符进行计数非常聪明。例如，如果我想做一个黎巴嫩国旗，我可以将两个Unicode字符组合在一起
Generally Swift is really smart about counting grapheme clusters as a single character. If I want to make a Lebanese flag, for example, I can combine the two Unicode characters
- U + 1F1F1区域指示符字母L
- U + 1F1E7区域指示符字母B
和预期的一样，这是Swift中的一个字符：
and as expected this is one character in Swift:
let s = "\u{1f1f1}\u{1f1e7}"
assert(s.characters.count == 1)
assert(s.utf16.count == 4)
assert(s.utf8.count == 8)
但是，假设我想制作一个Fitzpatrick Type-5的自行车表情符号。如果我合并
However, let's say I want to make a Bicyclist emoji of Fitzpatrick Type-5. If I combine
- U + 1F6B4自行车清单
- U + 1F3FE EMOJI MODIFIER FITZPATRICK TYPE-5
Swift将此组合计为两个字符！
Swift counts this combination as two characters!
let s = "\u{1f6b4}\u{1f3fe}"
assert(s.characters.count == 2) // <----- WHY?
assert(s.utf16.count == 4)
assert(s.utf8.count == 8)
这两个字符为什么不是一个？
Why is this two characters instead of one?
为了显示为什么我期望它是1，请注意，实际上该簇被解释为作为有效的表情符号：
To show why I would expect it be 1, note that this cluster is actually interpreted as a valid emoji:
推荐答案
部分答案在错误报告。将Unicode字符串拆分为字符时，Swift显然会使用 UAX＃中定义的Grapheme簇边界。 29 Unicode文本分段。有一个规则，不得在区域指示符之间切换，但没有这样的规则用于表情符号修饰符。因此，根据UAX＃29，字符串 \u {1f6b4} \u {1f3fe} 包含两个字素簇。请参阅Ken Whistler在Unicode邮件中的此消息列表进行解释：
Part of the answer is given in the bug report mentioned in emrys57's comment. When splitting a Unicode string into "characters", Swift apparently uses the Grapheme Cluster Boundaries defined in UAX #29 Unicode Text Segmentation. There's a rule not to break between regional indicator symbols, but there is no such rule for Emoji modifiers. So, according to UAX #29, the string "\u{1f6b4}\u{1f3fe}" contains two grapheme clusters. See this message from Ken Whistler on the Unicode mailing list for an explanation:
这是由于修饰符的后备行为是
而已，就像单独的象形图一样，即色板图像。 [...]您需要有关这些序列的其他特定
知识-这不仅仅是UAX＃29字素簇的
default 实现的结果。
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