"The ffi_ligature (U+FB03) is not decomposed, because it has a compatibility mapping, not a canonical mapping." Standard's text from Table 7 "what the standard seems to say is, that the ffi ligature is not equivalent to plain ffi because, if i write a string with the ligature, then it is a different string than plain "ffi", because both forms are printable. on the other hand ä and a" (two different encodings for ä) are the same, because the printed forms are always identical." --Martin I agree with that interpretation. I'm struggling to be clear about the consequences for equality testing. For sorting - every sort order can be created to go along with every encoding, so I suppose it just depends on which pre-assembled sort order is used. On 9 December 2015 at 02:36, Martin Bähr <mbaehr@email.archlab.tuwien.ac.at> wrote:

Excerpts from Martin Bähr's message of 2015-12-09 02:43:35 +0100:

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Excerpts from EuanM's message of 2015-12-09 01:59:43 +0100:

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http://www.unicode.org/reports/tr15/#Stable_Code_Points Table 7, the discussion of Ligatures, (which uses the ligature of "ffi" as its example)

ß is not a ligature of ss, but is a different character.

rereading this, i think i am wrong, in that this has nothing to do with ß vs ss.

looking at the standard i also don't understand your conclusion.

what the standard seems to say is, that the ffi ligature is not equivalent to plain ffi because, if i write a string with the ligature, then it is a different string than plain "ffi", because both forms are printable. on the other hand ä and a" (two different encodings for ä) are the same, because the printed forms are always identical.

however that doesn't mean that they are sorted differently. german sorting rules for example explicitly state that ß and ss are sorted the same. (at least according to wikipedia :-) and surely, ffi ligature and ffi are sorted the same too.

greetings, martin.

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