Question

我之前在Stack Overflow上发布了关于如何在c ++程序中旋转BMP图像的问题。但是，现在，我还有更多关于我的进展的信息。

我想知道在进行图像计算后，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。我应该提一下：

我不能将其他外部库用于此程序
这是一个小型图像处理程序，它有一个菜单系统

Answer 1

图像变换通常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。这样您就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stackoverflow中键入它，所以它充满了bug;尽管如此，这个概念仍然存在。

C ++方法中的图像旋转

1 个答案: