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LibGfx: Add Emoji::emoji_for_code_point_iterator(Utf8CodePointIterator&)
In the common case of text rendering rather than getting the emoji bitmap for a fixed number of code points, we don't know how many code points make one emoji. As far as I can tell, the longest ones are up to ten code points, so we try to consume all of them and do a lookup during each iteration, and return the emoji for the longest chain of code points. Quite basic and definitely room for improvement, but it works!
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@ -8,6 +8,7 @@
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#include <AK/HashMap.h>
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#include <AK/Span.h>
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#include <AK/String.h>
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#include <AK/Utf8View.h>
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#include <LibGfx/Bitmap.h>
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#include <LibGfx/Emoji.h>
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@ -41,4 +42,48 @@ Bitmap const* Emoji::emoji_for_code_points(Span<u32> const& code_points)
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return bitmap.ptr();
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}
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Bitmap const* Emoji::emoji_for_code_point_iterator(Utf8CodePointIterator& it)
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{
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// NOTE: I'm sure this could be more efficient, e.g. by checking if each code point falls
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// into a certain range in the loop below (emojis, modifiers, variation selectors, ZWJ),
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// and bailing out early if not. Current worst case is 10 file lookups for any sequence of
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// code points (if the first glyph isn't part of the font in regular text rendering).
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constexpr size_t max_emoji_code_point_sequence_length = 10;
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Vector<u32, max_emoji_code_point_sequence_length> code_points;
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struct EmojiAndCodePoints {
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Bitmap const* emoji;
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Span<u32> code_points;
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};
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Vector<EmojiAndCodePoints, max_emoji_code_point_sequence_length> possible_emojis;
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// Determine all existing emojis for the longest possible ZWJ emoji sequence,
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// or until we run out of code points in the iterator.
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for (size_t i = 0; i < max_emoji_code_point_sequence_length; ++i) {
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auto code_point = it.peek(i);
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if (!code_point.has_value())
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break;
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code_points.append(*code_point);
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if (auto const* emoji = emoji_for_code_points(code_points))
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possible_emojis.empend(emoji, code_points);
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}
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if (possible_emojis.is_empty())
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return nullptr;
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// If we found one or more matches, return the longest, i.e. last. For example:
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// U+1F3F3 - white flag
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// U+1F3F3 U+FE0F U+200D U+1F308 - rainbow flag
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auto& [emoji, emoji_code_points] = possible_emojis.last();
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// Advance the iterator, so it's on the last code point of our found emoji and
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// whoever is iterating will advance to the next new code point.
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for (size_t i = 0; i < emoji_code_points.size() - 1; ++i)
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++it;
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return emoji;
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}
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}
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