能用c++写个coons曲面吗

当然可以！下面是一个简单的 C++ 实现：

#include <iostream>
#include <vector>
#include <cmath>

using namespace std;

class Point {
public:
    double x, y, z;
    Point(double _x, double _y, double _z) : x(_x), y(_y), z(_z) {}
};

class CoonsPatch {
public:
    vector<Point> points;
    CoonsPatch(Point p0, Point p1, Point p2, Point p3) {
        // 计算边界曲线
        vector<Point> u0 = cubicBezier(p0, Point((p1.x - p0.x) / 3.0 + p0.x, (p1.y - p0.y) / 3.0 + p0.y, (p1.z - p0.z) / 3.0 + p0.z), Point((p1.x - p0.x) * 2.0 / 3.0 + p0.x, (p1.y - p0.y) * 2.0 / 3.0 + p0.y, (p1.z - p0.z) * 2.0 / 3.0 + p0.z), p1);
        vector<Point> u1 = cubicBezier(p2, Point((p3.x - p2.x) / 3.0 + p2.x, (p3.y - p2.y) / 3.0 + p2.y, (p3.z - p2.z) / 3.0 + p2.z), Point((p3.x - p2.x) * 2.0 / 3.0 + p2.x, (p3.y - p2.y) * 2.0 / 3.0 + p2.y, (p3.z - p2.z) * 2.0 / 3.0 + p2.z), p3);
        vector<Point> v0 = cubicBezier(p0, Point((p2.x - p0.x) / 3.0 + p0.x, (p2.y - p0.y) / 3.0 + p0.y, (p2.z - p0.z) / 3.0 + p0.z), Point((p2.x - p0.x) * 2.0 / 3.0 + p0.x, (p2.y - p0.y) * 2.0 / 3.0 + p0.y, (p2.z - p0.z) * 2.0 / 3.0 + p0.z), p2);
        vector<Point> v1 = cubicBezier(p1, Point((p3.x - p1.x) / 3.0 + p1.x, (p3.y - p1.y) / 3.0 + p1.y, (p3.z - p1.z) / 3.0 + p1.z), Point((p3.x - p1.x) * 2.0 / 3.0 + p1.x, (p3.y - p1.y) * 2.0 / 3.0 + p1.y, (p3.z - p1.z) * 2.0 / 3.0 + p1.z), p3);
        // 计算内部曲面
        for (int i = 0; i < u0.size(); i++) {
            for (int j = 0; j < v0.size(); j++) {
                double u = (double)i / (u0.size() - 1), v = (double)j / (v0.size() - 1);
                Point p = Point((1 - u) * u0[j].x + u * u1[j].x + (1 - v) * v0[i].x + v * v1[i].x - (1 - u) * (1 - v) * p0.x - (1 - u) * v * p1.x - u * (1 - v) * p2.x - u * v * p3.x,
                                (1 - u) * u0[j].y + u * u1[j].y + (1 - v) * v0[i].y + v * v1[i].y - (1 - u) * (1 - v) * p0.y - (1 - u) * v * p1.y - u * (1 - v) * p2.y - u * v * p3.y,
                                (1 - u) * u0[j].z + u * u1[j].z + (1 - v) * v0[i].z + v * v1[i].z - (1 - u) * (1 - v) * p0.z - (1 - u) * v * p1.z - u * (1 - v) * p2.z - u * v * p3.z);
                points.push_back(p);
            }
        }
    }
private:
    // 计算三次贝塞尔曲线
    vector<Point> cubicBezier(Point p0, Point p1, Point p2, Point p3) {
        vector<Point> points;
        for (int i = 0; i < 100; i++) {
            double t = (double)i / 99.0;
            double x = pow(1 - t, 3) * p0.x + 3.0 * pow(1 - t, 2) * t * p1.x + 3.0 * (1 - t) * pow(t, 2) * p2.x + pow(t, 3) * p3.x;
            double y = pow(1 - t, 3) * p0.y + 3.0 * pow(1 - t, 2) * t * p1.y + 3.0 * (1 - t) * pow(t, 2) * p2.y + pow(t, 3) * p3.y;
            double z = pow(1 - t, 3) * p0.z + 3.0 * pow(1 - t, 2) * t * p1.z + 3.0 * (1 - t) * pow(t, 2) * p2.z + pow(t, 3) * p3.z;
            Point p(x, y, z);
            points.push_back(p);
        }
        return points;
    }
};

int main() {
    // 构造一个简单的 Coons 曲面
    Point p0(0, 0, 0), p1(2, 0, 0), p2(0, 2, 0), p3(2, 2, 0);
    CoonsPatch coons(p0, p1, p2, p3);
    // 输出曲面上的点
    for (int i = 0; i < coons.points.size(); i++) {
        cout << "(" << coons.points[i].x << ", " << coons.points[i].y << ", " << coons.points[i].z << ")" << endl;
    }
    return 0;
}

该实现中，`Point` 表示一个三维点，`CoonsPatch` 表示一个 Coons 曲面。在 `CoonsPatch` 的构造函数中，我们首先计算出边界曲线（使用三次贝塞尔曲线），然后计算出内部曲面上的所有点。最后，我们可以通过遍历 `CoonsPatch` 的 `points` 成员来输出曲面上的所有点。