set关联容器去除重复地图点出现问题 error: no match for ‘operator＜’ (operand types are ‘const Point’ and ‘const Point

set中每个元素只包含一个关键字，set支持高效关键字查询---检查一个关键字是否在set中。

vector<sweep::MetricPoint2D> carpetPoint;
set<Point> gridPoint;void DealMap::onRobotPoseReceived(const geometry_msgs::PoseStampedPtr ptrRobotPose)
{if (robot_state->device_state.carpet_check == 1){sweep::MetricPoint2D tempPoint; //米制坐标点tempPoint.x = ptrRobotPose->pose.position.x;tempPoint.y = ptrRobotPose->pose.position.y;Point intPoint;intPoint.x = (int)(tempPoint.x / mapResolution);intPoint.y = (int)(tempPoint.y / mapResolution); if (gridPoint.find(intPoint) == gridPoint.end()){gridPoint.insert(intPoint);carpetPoint.push_back(tempPoint);}}
}

if调用返回一个迭代器，如果关键字在set中，迭代器指向该关键字，否则find返回尾后迭代器。

运行的时候出现错误

error: no match for ‘operator<’ (operand types are ‘const Point’ and ‘const Point’)

{ return __x < __y; }

原因分析：

if (gridPoint.find(intPoint) == gridPoint.end())条件执行会对key的大小作比较，作为自定义类型Point，本身无法做大小比较。

解决方法：使set的key能比较大小

自定义类型增加<操作符重载函数:

class Point
{public:int x;int y;Point(){x = 0;y = 0;}Point(int x, int y){this->x = x;this->y = y;}bool operator<(const Point& point){if(this->x < point.x){return true;}else if (this->x == point.x){if (this->y < point.y){return true;}}return false;}friend bool operator<(const Point&point1,const Point&point2){if(point1.x < point2.x){return true;}else if (point1.x == point2.x){if (point1.y < point2.y){return true;}}return false;}
};

set容器有其特定的排序准则，缺省情况下以operator<进行比较，也就是默认升序排列。