如果是STL容器,最小值算法返回迭代器,如果是數組,則返回地址。
下面是求的一個DPoint3d數組中距目標點最短距離的點:
struct distanceCriteria : std::binary_function<DPoint3d,DPoint3d,bool>
{
distanceCriteria(DPoint3d* targetPt):m_targetPt(*targetPt){}
bool operator()(const DPoint3d& pt1,const DPoint3d& pt2)
{
double dis1 = mdlVec_distanceXY((DVec3d*)&pt1,(DVec3d*)&m_targetPt);
double dis2 = mdlVec_distanceXY((DVec3d*)&pt2,(DVec3d*)&m_targetPt);
return ( dis1 < dis2 );
}
private:
DPoint3d m_targetPt;
};
DPoint3d* getShortest(DPoint3d* pArray,size_t ptNums,DPoint3d* targetPt)
{
return std::min_element(pArray, pArray + ptNums, distanceCriteria(targetPt));
}
如果採用C++ 11的lambda表達式,則會簡單很多,效果如同在函數裏定義函數,實際是定義了仿函數。DPoint3d* getShortest(DPoint3d* pArray, size_t ptNums, DPoint3d* targetPt)
{
return std::min_element(pArray,
pArray + ptNums,
[&](const DPoint3d& pt1, const DPoint3d& pt2)->bool
{
double dis1 = mdlVec_distanceXY((DVec3d*) &pt1,(DVec3d*)targetPt);
double dis2 = mdlVec_distanceXY((DVec3d*) &pt2,(DVec3d*)targetPt);
return (dis1 < dis2);
});
}