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Auto pair-removal #2940

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merged 5 commits into from
Jul 22, 2022
Merged

Auto pair-removal #2940

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55b69fd
auto pair-removal
Jul 2, 2022
e7e591e
autopairs removal: use doc autopairs
Jul 2, 2022
005f4c6
autopairs-removal: limit to one-char selections
Jul 2, 2022
41a8f43
use single_grapheme() to check if range is one char
Jul 10, 2022
9659bbc
fix errouneous deletes of " and other symmetric autopairs when at buf…
Jul 21, 2022
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26 changes: 20 additions & 6 deletions helix-term/src/commands.rs
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Original file line number Diff line number Diff line change
Expand Up @@ -2992,12 +2992,26 @@ pub mod insert {
(start, pos, None) // delete!
}
} else {
// delete char
(
graphemes::nth_prev_grapheme_boundary(text, pos, count),
pos,
None,
)
match (text.get_char(pos.saturating_sub(1)), text.get_char(pos)) {
(Some(_x), Some(_y)) if auto_pairs::DEFAULT_PAIRS.contains(&(_x, _y)) =>
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// delete both
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{
(
graphemes::nth_prev_grapheme_boundary(text, pos, count),
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This might get a bit tricky with the count. If we just delete n chars in both directions, this will be wrong in some cases. e.g. presently with this case:

hello, "["], goodbye("")

If you hit backspace 3 times, you get

hello[,] goodbye("")

like you'd expect. But if you use the command with a count, e.g. if you bound backspace to this in normal mode and then hit 3<backspace>, you end up with:

hello[g]oodbye("")

Though, I don't know how many people bind this in normal mode and use the count with it, so I don't know if I'd consider this a blocker. It might be tricky to get the logic right.

graphemes::nth_next_grapheme_boundary(text, pos, count),
None,
)
}
_ =>
// delete 1 char
{
(
graphemes::nth_prev_grapheme_boundary(text, pos, count),
pos,
None,
)
}
}
}
});
doc.apply(&transaction, view.id);
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