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Properly detect tiles with more than 4 colours
Fixes #1127, which was caused by a dumb logic error. Duh me.
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@@ -986,12 +986,16 @@ void process() {
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for (auto tile : png.visitAsTiles()) {
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ProtoPalette tileColors;
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AttrmapEntry &attrs = attrmap.emplace_back();
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uint8_t nbColorsInTile = 0;
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for (uint32_t y = 0; y < 8; ++y) {
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for (uint32_t x = 0; x < 8; ++x) {
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Rgba color = tile.pixel(x, y);
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if (!color.isTransparent()) { // Do not count transparency in for packing
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tileColors.add(color.cgbColor());
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// Add the color to the proto-pal (if not full), and count it if it was unique.
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if (tileColors.add(color.cgbColor())) {
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++nbColorsInTile;
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}
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}
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}
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}
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@@ -1033,9 +1037,9 @@ void process() {
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}
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}
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if (tileColors.size() > options.maxOpaqueColors()) {
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if (nbColorsInTile > options.maxOpaqueColors()) {
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fatal("Tile at (%" PRIu32 ", %" PRIu32 ") has %zu opaque colors, more than %" PRIu8 "!",
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tile.x, tile.y, tileColors.size(), options.maxOpaqueColors());
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tile.x, tile.y, nbColorsInTile, options.maxOpaqueColors());
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}
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attrs.protoPaletteID = protoPalettes.size();
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@@ -17,24 +17,29 @@
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bool ProtoPalette::add(uint16_t color) {
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size_t i = 0;
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// Seek the first slot greater than our color
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// Seek the first slot greater than the new color
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// (A linear search is better because we don't store the array size,
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// and there are very few slots anyway)
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while (_colorIndices[i] < color) {
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++i;
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if (i == _colorIndices.size())
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return false;
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if (i == _colorIndices.size()) {
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// We reached the end of the array without finding the color, so it's a new one.
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return true;
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}
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}
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// If we found it, great! Nothing else to do.
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if (_colorIndices[i] == color) {
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return false;
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}
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// If we found ourselves, great!
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if (_colorIndices[i] == color)
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return true;
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// Swap entries until the end
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while (_colorIndices[i] != UINT16_MAX) {
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std::swap(_colorIndices[i], color);
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++i;
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if (i == _colorIndices.size())
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return false; // Oh well
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if (i == _colorIndices.size()) {
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// The set is full, but doesn't include the new color.
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return true;
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}
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}
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// Write that last one into the new slot
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_colorIndices[i] = color;
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