C++方法中的图像旋转

Image Rotation in a C++ Method

本文关键字：图像 旋转 方法 C++      更新时间：2023-10-16

我之前在Stack Overflow上发布了有关如何在c ++程序中精确旋转BMP图像的问题。然而，现在，关于我的进步，我有更多的东西要展示。

我

想知道在我进行图像计算后，我的程序如何（或为什么）不会输出图像：

void BMPImage::Rotate45Left(float point1, float point2, float point3)
{
  float radians = (2 * 3.1416*45) / 360;
  float cosine = (float)cos(radians);
  float sine = (float)sin(radians);
  float point1Xtreme = 0;
  float point1Yearly = 0;
  float point2Xtreme = 0;
  float point2Yearly = 0;
  float point3Xtreme = 0;
  float point3Yearly = 0;
int SourceBitmapHeight = m_BIH.biHeight;
int SourceBitmapWidth = m_BIH.biWidth;
point1Xtreme = (-m_BIH.biHeight*sine);
point1Yearly = (m_BIH.biHeight*cosine);
point2Xtreme = (m_BIH.biWidth*cosine - m_BIH.biHeight*sine);
point2Yearly = (m_BIH.biHeight*cosine + m_BIH.biWidth*sine);
point3Xtreme = (m_BIH.biWidth*cosine);
point3Yearly = (m_BIH.biWidth*sine);
float Minx = min(0, min(point1Xtreme, min(point2Xtreme, point3Xtreme)));
float Miny = min(0, min(point1Yearly, min(point2Yearly, point3Yearly)));
float Maxx = max(point1Xtreme, max(point2Xtreme, point3Xtreme));
float Maxy = max(point1Yearly, max(point2Yearly, point3Yearly));
int FinalBitmapWidth = (int)ceil(fabs(Maxx) - Minx);
int FinalBitmapHeight = (int)ceil(fabs(Maxy) - Miny);
FinalBitmapHeight = m_BIH.biHeight;
FinalBitmapWidth = m_BIH.biWidth;
int finalBitmap;

如果有人有任何有用的指示，那就太好了。我应该提到：

  • 我不能为这个程序的目的使用其他外部库
  • 它是一个小型图像处理程序，具有菜单系统

图像转换通常是通过将目标像素投影到源像素上，然后计算该目标像素的值来完成的。这样，您可以轻松地合并不同的插值方法。

template <typename T>
struct Image {
    Image(T* data, size_t rows, size_t cols) : 
        data_(data), rows_(rows), cols_(cols) {}
    T* data_;
    size_t rows_;
    size_t cols_;
    T& operator()(size_t row, size_t col) {
        return data_[col + row * cols_];
    }
 };
template <typename T>
T clamp(T value, T lower_bound, T upper_bound) {
    value = std::min(std::max(value, lower_bound), upper_bound);
}
void rotate_image(Image const &src, Image &dst, float ang) {
    // Affine transformation matrix 
    // H = [a, b, c]
    //     [d, e, f]
    // Remember, we are transforming from destination to source, 
    // thus the negated angle. 
    float H[] = {cos(-ang), -sin(-ang), dst.cols_/2 - src.cols_*cos(-ang)/2,
                 sin(-ang),  cos(-ang), dst.rows_/2 - src.rows_*cos(-ang)/2}; 

    for (size_t row = 0; row < dst.rows_; ++row) {
       for (size_t col = 0; col < dst.cols_; ++cols) {
           int src_col = round(H[0] * col + H[1] * row + H[2]);
           src_col = clamp(src_col, 0, src.cols_ - 1);
           int src_row = round(H[3] * col + H[4] * row + H[5]);
           src_row = clamp(src_row, 0, src.rows_ - 1);               
           dst(row, col) = src(src_row, src_col);
       }
    }
}

上述方法以任意角度旋转图像并使用最近邻插值。我直接在堆栈溢出中键入它，所以它充满了错误;尽管如此，这个概念还是存在的。

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