Layer type evaluators #9064
Layer type evaluators #9064
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@kkaefer Do the changes to |
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| Layer* operator()(style::LineLayer&) && { return new LineLayer(map, cast<style::LineLayer>()); } | ||
| Layer* operator()(style::RasterLayer&) && { return new RasterLayer(map, cast<style::RasterLayer>()); } | ||
| Layer* operator()(style::SymbolLayer&) && { return new SymbolLayer(map, cast<style::SymbolLayer>()); } | ||
| Layer* operator()(style::CustomLayer&) && { return new CustomLayer(map, cast<style::CustomLayer>()); } |
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This could be a template function?
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@kkaefer I was hoping the same, but don't quite know how. The classes involved are not the same type, but named the same, eg mbgl::style::LineLayer and mbgl::android::LineLayer. Where the latter is a Peer class wrapping the former.
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This could be done by adding a typedef member to the mbgl::android::LineLayer with a shared name (e.g. using LayerType = style::LineLayer in mbgl::android::LineLayer, then using a function like
template <class PeerType>
Layer* operator(typename PeerType::LayerType&) && {
return new PeerType(map, cast<typename PeerType::LayerType>());
}
| Layer* operator()(style::LineLayer& layer) { return new LineLayer(map, layer); } | ||
| Layer* operator()(style::RasterLayer& layer) { return new RasterLayer(map, layer); } | ||
| Layer* operator()(style::SymbolLayer& layer) { return new SymbolLayer(map, layer); } | ||
| Layer* operator()(style::CustomLayer& layer) { return new CustomLayer(map, layer); } |
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Could this be a template function?
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Closing as this PR won't trigger CI. Replaced by #9069 |
Fixes #8758