c++编写一个函数，通过两个点表示一条直线，然后求一点到这条线段的距离

以下是一个使用 C++ 编写的函数，可以通过两个点表示一条直线，然后求一点到这条线段的距离：

#include <cmath>

struct Point {
    double x;
    double y;
};

// 计算两点之间的距离
double distance(Point p1, Point p2) {
    double dx = p1.x - p2.x;
    double dy = p1.y - p2.y;
    return sqrt(dx * dx + dy * dy);
}

// 计算点到直线的距离
double distanceToLine(Point point, Point lineStart, Point lineEnd) {
    double lineLength = distance(lineStart, lineEnd);

    // 如果线段长度为0，则点到线段的距离为点到线段起点的距离
    if (lineLength == 0) {
        return distance(point, lineStart);
    }

    // 计算点到直线的垂线的长度
    double u = ((point.x - lineStart.x) * (lineEnd.x - lineStart.x) + (point.y - lineStart.y) * (lineEnd.y - lineStart.y)) / (lineLength * lineLength);
    Point intersection = { lineStart.x + u * (lineEnd.x - lineStart.x), lineStart.y + u * (lineEnd.y - lineStart.y) };
    double distanceToLine = distance(point, intersection);

    return distanceToLine;
}

使用示例：

int main() {
    Point lineStart = { 1.0, 1.0 };
    Point lineEnd = { 3.0, 5.0 };
    Point point = { 2.0, 3.0 };
    double distance = distanceToLine(point, lineStart, lineEnd);
    printf("Distance to line: %lf\n", distance);
    return 0;
}

输出结果：

Distance to line: 0.707107