After taking a long hard look at the specification for computing the boundaries for extended grapheme clusters (EGCs) at https://www.unicode.org/reports/tr29/#Grapheme_Cluster_Boundary_Rules,
it is obvious that the rules for EGCs all have the shape of describing when it is allowed to append a code point to an existing EGC to form a longer EGC. From that fact alone my two questions follow: 1) Yes, every non-empty prefix of code points which form an EGC is also an EGC. 2) No, by adding a code point to a valid Unicode string you will not decrease its length in terms of number of EGCs it consists of.
So, given this, the following Swift code will extract the longest Unicode character from the start of a byte sequence (or return nil if there is no valid Unicode character there):
func lex<S : Sequence>(_ input : S) -> (length : Int, out: Character)? where S.Element == UInt8 {
// This code works under three assumptions, all of which are true:
// 1) If a sequence of codepoints does not form a valid character, then appending codepoints to it does not yield a valid character
// 2) Appending codepoints to a sequence of codepoints does not decrease its length in terms of extended grapheme clusters
// 3) a codepoint takes up at most 4 bytes in an UTF8 encoding
var chars : [UInt8] = []
var result : String = ""
var resultLength = 0
func value() -> (length : Int, out : Character)? {
guard let character = result.first else { return nil }
return (length: resultLength, out: character)
}
var length = 0
var iterator = input.makeIterator()
while length - resultLength <= 4 {
guard let char = iterator.next() else { return value() }
chars.append(char)
length += 1
guard let s = String(bytes: chars, encoding: .utf8) else { continue }
guard s.count == 1 else { return value() }
result = s
resultLength = length
}
return value()
}
init?(validatingUTF8:)to check incrementally longer prefixes isn't a bad idea. While extended grapheme clusters can get pathologically big (try copying this answer into a unicode inspector lol), the vast majority of them are tiny.findLongestValidCharacter(in input: [UnicodeScalar]) -> Character. Even if such a function existed, Pathologically long characters would still be an issue. The "bad" cases would be less bad given that the function could beO(input.count)rather than theO(input.count^2)of the brute force approach, but that can still get bad.