Further simplify tile subdivision mechanism#2525
Further simplify tile subdivision mechanism#2525jailln wants to merge 7 commits intoiTowns:masterfrom
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jailln
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I agree that pendingSubdivision should be removed. At the time of development, we introduced tile asynchronicity to handle streamed DEM tiles or to compute tiles in workers. |
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We need to remove |
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When you remove tiles instancing from scheduler, there is no more earlyDropFunction mechanism, to avoid unnecessary instancing. |
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Sorry, |
| } catch (e) { | ||
| return Promise.reject(e); | ||
| } | ||
| // TODO removed a try catch here that didn't seem needed after a first quick look, but needs to be re-checked |
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Yes, there's no need to add a try block. Could remove comment
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| function _minimizeNetworkTraffic(node, nodeLevel, currentLevel, source) { | ||
| // TO DO source.isVectorTileSource is a temp fix for pendingSubdivision. | ||
| // see issue https://github.com/iTowns/itowns/issues/2214 | ||
| if (node.pendingSubdivision && !source.isVectorTileSource) { | ||
| return currentLevel; | ||
| } | ||
| function _minimizeNetworkTraffic(node, nodeLevel) { | ||
| return nodeLevel; | ||
| } |
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this method could be removed
| case STRATEGY_MIN_NETWORK_TRAFFIC: | ||
| default: | ||
| nextLevelToFetch = _minimizeNetworkTraffic(node, nodeLevel, currentLevel, layer.source); | ||
| nextLevelToFetch = _minimizeNetworkTraffic(node, nodeLevel); |
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If you remove _minimizeNetworkTraffic
| nextLevelToFetch = _minimizeNetworkTraffic(node, nodeLevel); | |
| nextLevelToFetch = nodeLevel |
| if (node.parent.pendingSubdivision) { | ||
| node.visible = false; | ||
| node.material.visible = false; | ||
| this.info.update(node); |
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the updating info is released in each case ?
| // display iff children aren't ready | ||
| node.material.visible = node.pendingSubdivision; | ||
| node.material.visible = false; | ||
| this.info.update(node); |
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this.info.update(node); Could this call be written only once in this method?
| */ | ||
| subdivideNode(context, node) { | ||
| if (!node.pendingSubdivision && !node.children.some(n => n.layer == this)) { | ||
| if (!node.children.some(n => n.layer == this)) { |
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may be could be write
if (node.children.length)Because the children can only be on the same node's layer (to be verified).
3 participants
Based on #2521
Related to #2522