[求助]：求2条线交点。。。。。。。

梦幻神话

求助：
提供2条线，如何判断它们是否有交点，如果有如何求？
谢谢。

jacking

p->intersectWith(AcDbEntity* pEnt,AcDb::Intersect intType,AcGePoint3dArray& points, 0, 0) ;
其中有个是否延伸的参数AcDb::Intersect intType 取值分别如下：
1、AcDb::kOnBothOperands
Do not extend either entity. This results in only calculating intersections where the two entitys' geometry actually intersect
2、AcDb::kExtendThis
Extend this entity (if necessary) when calculating intersections, but do not extend the pEnt entity.
3、AcDb::kExtendArg
Extend the pEnt entity (if necessary) when calculating intersections, but do not extend this entity.
4、AcDb::kExtendBoth
Extend both the pEnt entity and this entity (if necessary) when calculating intersections

如果取得交点，放入points数组中，points的长度即为交点个数，用points[0],points[1]……得到各交点

梦幻神话

谢谢，有收获。
不过函数
virtual Acad::ErrorStatus  intersectWith(const AcDbEntity* pEnt, AcDb::Intersect intType, const AcGePlane& projPlane, AcGePoint3dArray& points, int thisGsMarker = 0,int otherGsMarker = 0) const
只是求pEnt与其他的实体的交点，我现在只是想求指定的2条线的焦点而已，不知道如何解决。。。

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理解错误，刚才头脑发热，没看清楚。。。
现在理解了。。。

小树袋熊

线也是实体啊

ishou

int

acdbInters(

const ads_point from1,

const ads_point to1,

const ads_point from2,

const ads_point to2,

int teston,

ads_point result);

from1 The beginning point of the first line
to1 The end point of the first line
from2 The beginning point of the second line
to2 The end point of the second line
teston If nonzero, the intersection must lie on both of the specified segments; if zero, the intesection can be off one or both segments
result The point of intersection

Examines two lines and returns the point where they intersect.

All four point arguments are three-dimensional points; acdbInters() checks for intersection in 3D space. All points should be expressed in terms of the same coordinate system. Otherwise, acdbInters() will return garbage (it calculates a result but does not check the input values against the current drawing).

If acdbInters() finds an intersection point, it returns RTNORM; otherwise, it returns an error code.

cxm1212

可以调用IntersectWith函数，函数带有一个对象为AcGe3dPointArray的参数，通过判断数组是否为空来判断两条线是否有交点