Does Boost geometry have line and line segment intersection?
0
I have seen example of line segment(s1) and line segment(s2) example but I have not came across line (l) and line segment(s) intersection. Does any one know where I can find one or if there is any.
I tried to use linestring (l) and segment (s) but I get error when I try to use them as
typedef boost::geometry::model::point<double, 2, boost::geometry::cs::cartesian> boostPoint;
std::vector<boostPoint> output;
boost::geometry::intersection(line, seg,output);
c++ boost geometry boost-geometry
asked Nov 20 at 19:00
mato
667
add a comment |
0
I have seen example of line segment(s1) and line segment(s2) example but I have not came across line (l) and line segment(s) intersection. Does any one know where I can find one or if there is any.
I tried to use linestring (l) and segment (s) but I get error when I try to use them as
typedef boost::geometry::model::point<double, 2, boost::geometry::cs::cartesian> boostPoint;
std::vector<boostPoint> output;
boost::geometry::intersection(line, seg,output);
c++ boost geometry boost-geometry
asked Nov 20 at 19:00
mato
667
Yes,intersectionforsegmentandlinestringdoesn't compile, but we know that any segment can be represented bylinestringso you can useintersectionfor two linestrings objects, in this case code compiles. I know that the use of linestring to represent segment is redundant (segmentuses pair of points,linestringuses vector to keeps points) but code compiles and works, so use linestrings.
– rafix07
Nov 20 at 19:39
But aren't line (line string) and line segment (segment) fundamentally different. Line extend to infinite and segment does not.
– mato
Nov 20 at 20:53
In Boost segment is finite line which contains 2 points (start point and end point), it is not considered as infinite line while calculating the interesection points. So I think you cannot do the calculations you want (segments treated as inifinite lines).
– rafix07
Nov 20 at 21:02
@rafix07 it depends on the actual goal (which is not specified in the question). There is, e.g. a distance strategy that uses "projected point" which does use the "extended line" for a segment
– sehe
Nov 20 at 23:58
1
@sehe Yes, you are right as to projected point strategy, mabye it would be useful in this case, but OP's question should be more clear.
– rafix07
Nov 21 at 7:39
add a comment |
0
0
0
I have seen example of line segment(s1) and line segment(s2) example but I have not came across line (l) and line segment(s) intersection. Does any one know where I can find one or if there is any.
I tried to use linestring (l) and segment (s) but I get error when I try to use them as
typedef boost::geometry::model::point<double, 2, boost::geometry::cs::cartesian> boostPoint;
std::vector<boostPoint> output;
boost::geometry::intersection(line, seg,output);
c++ boost geometry boost-geometry
asked Nov 20 at 19:00
mato
667
I have seen example of line segment(s1) and line segment(s2) example but I have not came across line (l) and line segment(s) intersection. Does any one know where I can find one or if there is any.
I tried to use linestring (l) and segment (s) but I get error when I try to use them as
typedef boost::geometry::model::point<double, 2, boost::geometry::cs::cartesian> boostPoint;
std::vector<boostPoint> output;
boost::geometry::intersection(line, seg,output);
c++ boost geometry boost-geometry
c++ boost geometry boost-geometry
asked Nov 20 at 19:00
mato
667
asked Nov 20 at 19:00
mato
667
edited Nov 20 at 19:26
asked Nov 20 at 19:00
mato
667
asked Nov 20 at 19:00
mato
667
asked Nov 20 at 19:00
mato
667
667
Yes,intersectionforsegmentandlinestringdoesn't compile, but we know that any segment can be represented bylinestringso you can useintersectionfor two linestrings objects, in this case code compiles. I know that the use of linestring to represent segment is redundant (segmentuses pair of points,linestringuses vector to keeps points) but code compiles and works, so use linestrings.
– rafix07
Nov 20 at 19:39
But aren't line (line string) and line segment (segment) fundamentally different. Line extend to infinite and segment does not.
– mato
Nov 20 at 20:53
In Boost segment is finite line which contains 2 points (start point and end point), it is not considered as infinite line while calculating the interesection points. So I think you cannot do the calculations you want (segments treated as inifinite lines).
– rafix07
Nov 20 at 21:02
@rafix07 it depends on the actual goal (which is not specified in the question). There is, e.g. a distance strategy that uses "projected point" which does use the "extended line" for a segment
– sehe
Nov 20 at 23:58
1
@sehe Yes, you are right as to projected point strategy, mabye it would be useful in this case, but OP's question should be more clear.
– rafix07
Nov 21 at 7:39
add a comment |
Yes,intersectionforsegmentandlinestringdoesn't compile, but we know that any segment can be represented bylinestringso you can useintersectionfor two linestrings objects, in this case code compiles. I know that the use of linestring to represent segment is redundant (segmentuses pair of points,linestringuses vector to keeps points) but code compiles and works, so use linestrings.
– rafix07
Nov 20 at 19:39
But aren't line (line string) and line segment (segment) fundamentally different. Line extend to infinite and segment does not.
– mato
Nov 20 at 20:53
In Boost segment is finite line which contains 2 points (start point and end point), it is not considered as infinite line while calculating the interesection points. So I think you cannot do the calculations you want (segments treated as inifinite lines).
– rafix07
Nov 20 at 21:02
@rafix07 it depends on the actual goal (which is not specified in the question). There is, e.g. a distance strategy that uses "projected point" which does use the "extended line" for a segment
– sehe
Nov 20 at 23:58
1
@sehe Yes, you are right as to projected point strategy, mabye it would be useful in this case, but OP's question should be more clear.
– rafix07
Nov 21 at 7:39
Yes,
intersection for segment and linestring doesn't compile, but we know that any segment can be represented by linestring so you can use intersection for two linestrings objects, in this case code compiles. I know that the use of linestring to represent segment is redundant (segment uses pair of points, linestring uses vector to keeps points) but code compiles and works, so use linestrings.– rafix07
Nov 20 at 19:39
Yes,
intersection for segment and linestring doesn't compile, but we know that any segment can be represented by linestring so you can use intersection for two linestrings objects, in this case code compiles. I know that the use of linestring to represent segment is redundant (segment uses pair of points, linestring uses vector to keeps points) but code compiles and works, so use linestrings.– rafix07
Nov 20 at 19:39
But aren't line (line string) and line segment (segment) fundamentally different. Line extend to infinite and segment does not.
– mato
Nov 20 at 20:53
But aren't line (line string) and line segment (segment) fundamentally different. Line extend to infinite and segment does not.
– mato
Nov 20 at 20:53
In Boost segment is finite line which contains 2 points (start point and end point), it is not considered as infinite line while calculating the interesection points. So I think you cannot do the calculations you want (segments treated as inifinite lines).
– rafix07
Nov 20 at 21:02
In Boost segment is finite line which contains 2 points (start point and end point), it is not considered as infinite line while calculating the interesection points. So I think you cannot do the calculations you want (segments treated as inifinite lines).
– rafix07
Nov 20 at 21:02
@rafix07 it depends on the actual goal (which is not specified in the question). There is, e.g. a distance strategy that uses "projected point" which does use the "extended line" for a segment
– sehe
Nov 20 at 23:58
@rafix07 it depends on the actual goal (which is not specified in the question). There is, e.g. a distance strategy that uses "projected point" which does use the "extended line" for a segment
– sehe
Nov 20 at 23:58
1
1
@sehe Yes, you are right as to projected point strategy, mabye it would be useful in this case, but OP's question should be more clear.
– rafix07
Nov 21 at 7:39
@sehe Yes, you are right as to projected point strategy, mabye it would be useful in this case, but OP's question should be more clear.
– rafix07
Nov 21 at 7:39
add a comment |
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Yes,
intersectionforsegmentandlinestringdoesn't compile, but we know that any segment can be represented bylinestringso you can useintersectionfor two linestrings objects, in this case code compiles. I know that the use of linestring to represent segment is redundant (segmentuses pair of points,linestringuses vector to keeps points) but code compiles and works, so use linestrings.– rafix07
Nov 20 at 19:39
But aren't line (line string) and line segment (segment) fundamentally different. Line extend to infinite and segment does not.
– mato
Nov 20 at 20:53
In Boost segment is finite line which contains 2 points (start point and end point), it is not considered as infinite line while calculating the interesection points. So I think you cannot do the calculations you want (segments treated as inifinite lines).
– rafix07
Nov 20 at 21:02
@rafix07 it depends on the actual goal (which is not specified in the question). There is, e.g. a distance strategy that uses "projected point" which does use the "extended line" for a segment
– sehe
Nov 20 at 23:58
1
@sehe Yes, you are right as to projected point strategy, mabye it would be useful in this case, but OP's question should be more clear.
– rafix07
Nov 21 at 7:39