LineSegmenter returns breakpoint for Combining Marks

[saecki]

For the following example the LineSegmenter returns breakpoints which are classified as LineBreak::CombiningMark by icu_properties.

// [dependencies]
// icu_segmenter = "1.5.0"
// icu_properties = "1.5.1"

fn main() {
    let segmenter = icu_segmenter::LineSegmenter::new_auto();
    let linebreaks = icu_properties::maps::line_break();

    let str = "\u{2066}\u{2069}";
    let iter = segmenter.segment_str(str);
    for brk_idx in iter {
        let Some(c) = str[..brk_idx].chars().next_back() else {
            continue;
        };
        let lb = linebreaks.get(c);
        println!("{brk_idx:3}: {lb:?} {}", c.escape_unicode());
    }
}

Produces this output:

  3: LineBreak(9) \u{2066}
  9: LineBreak(9) \u{2069}

The docs say:

The segmenter returns mandatory breaks (as defined by definition LD7 of Unicode Standard Annex #14, Unicode Line Breaking Algorithm) as well as line break opportunities (definition LD3).

Is this an oversight or is the returned breakpoint to be expected?

Relevant issue typst/typst#5489

[saecki]

Seems like this is correct behavior. See typst/typst#5489.