在正方形上使用时的 C++ 旋转矩阵问题

Question

我正在尝试在 C++ 中应用一个旋转矩阵，它围绕指定的原点将正方形的所有点旋转指定的度数。问题在于它基于 win32 控制台，因此每个点都必须对应一对整数，而不是浮点值。如下图所示，旋转正方形的整体形状与预期结果一致，但其中存在多个“孔”。

这是我的源代码:

#include <iostream>
#include <cmath>

using namespace std;

enum {W = 50, H = 50, S = 25}; //Width, Height, Square size

struct Vector2i
{
    int x;
    int y;
    Vector2i() {}
    Vector2i(int _x, int _y) : x(_x), y(_y) {}
};

struct Square
{
    bool Data[W][H];
    Vector2i Origin = Vector2i(W / 2, H / 2);

    void clear() {
        for (int y = 0; y < H; ++y) {
            for (int x = 0; x < W; ++x)
                Data[x][y] = false;
        }
    }

    void setSquare() {
        for (int y = H / 2 - S / 2; y < H / 2 + S / 2; ++y) {
            for (int x = W / 2 - S / 2; x < W / 2 + S / 2; ++x)
                Data[x][y] = true;
        }
    }

    void draw() {
        for (int y = 0; y < H; ++y) {
            for (int x = 0; x < W; ++x) {
                if (y == Origin.y && x == Origin.x) std::cout << '+'; //Marks the origin
                else if (Data[x][y]) std::cout << 'X';
                else std::cout << '.';
            }
            std::cout << '\n';
        }
    }
};

Vector2i newPos(Vector2i old, double theta) {
    theta *= 3.14159265d / 180.d; //Converting from degrees to radians

    int X = ceil(cos(theta) * old.x - sin(theta) * old.y);
    int Y = ceil(sin(theta) * old.x + cos(theta) * old.y);

    return Vector2i(X, Y);
}

int main()
{
    cout << "Enter an angle (in degrees): ";
    double angle = 0;
    cin >> angle;

    Square One;
    One.clear();
    One.setSquare();
    One.draw();

    Square Two;
    Two.clear();

    ///Draw the rotated square as the second square
    for (int y = 0; y < H; ++y) {
        for (int x = 0; x < W; ++x) {
            if (One.Data[x][y]) {
                Vector2i finalVec = newPos(Vector2i(x - One.Origin.x, 
                                                    y - One.Origin.y), angle);
                Two.Data[finalVec.x + One.Origin.x][finalVec.y + One.Origin.y] = true;
            }
        }
    }

    ///Copy the second square back into the first
    for (int y = 0; y < H; ++y) {
        for (int x = 0; x < W; ++x)
            One.Data[x][y] = Two.Data[x][y];
    }

    One.draw();

    return 0;
}

这是由于 newPos() 函数的准确性造成的，还是由于舍入到 int 值导致的？

此外，有没有办法解决这个问题或预测漏洞的位置？

编辑:已解决!

根据 infgeoax 向后工作的建议，我创建了一个函数来计算原始位置。我会把增强的代码留在这里，以防将来有人遇到类似的问题(感谢大家的帮助![特别是 infgeoax，用于脑波]):

#include <iostream>
#include <cmath>

using namespace std;

enum {W = 50, H = 50, S = 25};

struct Vector2i
{
    int x;
    int y;
    Vector2i() {}
    Vector2i(int _x, int _y) : x(_x), y(_y) {}
};

struct Square
{
    bool Data[W][H];;
    Vector2i Origin = Vector2i(W / 2, H / 2);

    void clear() {
        for (int y = 0; y < H; ++y) {
            for (int x = 0; x < W; ++x)
                Data[x][y] = false;
        }
    }

    void setSquare() {
        for (int y = H / 2 - S / 2; y < H / 2 + S / 2; ++y) {
            for (int x = W / 2 - S / 2; x < W / 2 + S / 2; ++x)
                Data[x][y] = true;
        }
    }

    void draw() {
        for (int y = 0; y < H; ++y) {
            for (int x = 0; x < W; ++x) {
                if (y == Origin.y && x == Origin.x) std::cout << '+'; //Marks the origin
                else if (Data[x][y]) std::cout << 'X';
                else std::cout << '.';
            }
            std::cout << '\n';
        }
    }
};

Vector2i oldPos(Vector2i new_, float theta) {
    theta *= 3.14159265f / 180.f; //Converting from degrees to radians
    return Vector2i(new_.x * cosf(theta) + new_.y * sinf(theta) + 0.5f,
                    new_.y * cosf(theta) - new_.x * sinf(theta) + 0.5f);
}

int main()
{
    cout << "Enter an angle (in degrees): ";
    float angle = 0;
    cin >> angle;

    Square One;
    One.clear();
    One.setSquare();
    One.draw();

    Square Two;
    Two.clear();

    for (int y = 0; y < H; ++y) {
        for (int x = 0; x < W; ++x) {
            Vector2i vec = oldPos(Vector2i(x - One.Origin.x, y - One.Origin.y), angle);
            vec.x += One.Origin.x;
            vec.y += One.Origin.y;
            if (vec.x >= 0 && vec.x < W && vec.y >= 0 && vec.y < H)
                Two.Data[x][y] = One.Data[vec.x][vec.y];
        }
    }

    Two.draw();

    return 0;
}

Answer 1

您的问题与您开发的是控制台应用程序还是 GUI 应用程序无关。图像作为像素矩阵存储和处理。旋转图像时，特定像素的结果位置通常不是整数。

想法是反过来。

1. 计算旋转正方形的四个角。
2. 对于旋转正方形中的每个位置(像素)，通过将其旋转回原始正方形来计算其颜色。