SelectEdgeByProjectedCoordinate #11
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…or which the specified coordinate is closest to its perpendicular projection onto this edge.
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What you're caclulating is not necessarily the distance to the projected point. If the projection lies outside of a line segment, the distance to the closest endpoint of the line segment (edge) is computed. The naming is somewhat misleading and probably unnecessarily complex, I guess. Why not simply |
tools/selection_tools.cpp
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| maxDeviationAngle, selectFlipped); | ||
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| Edge* SelectEdgeByMinPerpendicularDistance( |
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As commented somewhere, this is not accurate - if the projection of the given coordinate on an edge lies outside of that edge, the distance to the closest endpoint is considered.
tools/selection_tools.cpp
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| return NULL; | ||
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| EdgeIterator iter = grid.begin<Edge>(); | ||
| Edge* bestEdge = *iter; |
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maybe initialize with NULL and add a check below inside the loop:
if (bestEdge == NULL || dist < bestDist)
{...}
You could then use a for loop and avoid code duplication. You could also remove the empty-check above.
…ClosestEdge' and streamlining the code according to the suggestions made.
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Dear Sebastian, thank you for your review -- indeed, the naming was somewhat misleading. I have applied modifications according to your suggestions streamlining the code. Best regards, |
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Dear Martin, I'm sorry that it took so long to merge your request. Thanks for taking the time to consider my comments :) Best, |
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Added function 'SelectEdgeByProjectedCoordinate' selecting the edge for which the specified coordinate is closest to its perpendicular projection onto this edge.