Question

我正在尝试使用我发现here的双线性技术重新调整图像大小，但除了黑色图像外我什么都看不到。所以，首先我用LodePNG解码我的图像，像素进入vector<unsigned char>变量。它说它们存储为RGBARGBA但是当我尝试将图像应用到X11窗口时，我意识到它们存储为BGRABGRA。我不知道是否是更改顺序的X11 API或LodePNG解码器。无论如何，在任何事情之前，我将BGR转换为RGB：

// Here is where I have the pixels stored
vector<unsigned char> Image;

// Converting BGRA to RGBA, or vice-versa, I don't know, but it's how it is shown
// correctly on the window
unsigned char red, blue;
unsigned int i;
for(i=0; i<Image.size(); i+=4)
{
    red  = Image[i + 2];
    blue = Image[i];
    Image[i] = red;
    Image[i + 2] = blue;
}

所以，现在我尝试在将图像应用到窗口之前更改图像的大小。大小将是窗口的大小（拉伸它）。我首先尝试将RGBA转换为int值，如下所示：

vector<int> IntImage;
for(unsigned i=0; i<Image.size(); i+=4)
{
    IData.push_back(256*256*this->Data[i+2] + 256*this->Data[i+1] + this->Data[i]);
}

现在我从上面指定的链接中获得了这个函数，它应该进行插值：

vector<int> resizeBilinear(vector<int> pixels, int w, int h, int w2, int h2) {
    vector<int> temp(w2 * h2);
    int a, b, c, d, x, y, index ;
    float x_ratio = ((float)(w-1))/w2 ;
    float y_ratio = ((float)(h-1))/h2 ;
    float x_diff, y_diff, blue, red, green ;

    for (int i=0;i<h2;i++) {
        for (int j=0;j<w2;j++) {
            x = (int)(x_ratio * j) ;
            y = (int)(y_ratio * i) ;
            x_diff = (x_ratio * j) - x ;
            y_diff = (y_ratio * i) - y ;
            index = (y*w+x) ;                
            a = pixels[index] ;
            b = pixels[index+1] ;
            c = pixels[index+w] ;
            d = pixels[index+w+1] ;

            // blue element
            // Yb = Ab(1-w)(1-h) + Bb(w)(1-h) + Cb(h)(1-w) + Db(wh)
            blue = (a&0xff)*(1-x_diff)*(1-y_diff) + (b&0xff)*(x_diff)*(1-y_diff) +
                   (c&0xff)*(y_diff)*(1-x_diff)   + (d&0xff)*(x_diff*y_diff);

            // green element
            // Yg = Ag(1-w)(1-h) + Bg(w)(1-h) + Cg(h)(1-w) + Dg(wh)
            green = ((a>>8)&0xff)*(1-x_diff)*(1-y_diff) + ((b>>8)&0xff)*(x_diff)*(1-y_diff) +
                    ((c>>8)&0xff)*(y_diff)*(1-x_diff)   + ((d>>8)&0xff)*(x_diff*y_diff);

            // red element
            // Yr = Ar(1-w)(1-h) + Br(w)(1-h) + Cr(h)(1-w) + Dr(wh)
            red = ((a>>16)&0xff)*(1-x_diff)*(1-y_diff) + ((b>>16)&0xff)*(x_diff)*(1-y_diff) +
                  ((c>>16)&0xff)*(y_diff)*(1-x_diff)   + ((d>>16)&0xff)*(x_diff*y_diff);

            temp.push_back( 
                    ((((int)red)<<16)&0xff0000) |
                    ((((int)green)<<8)&0xff00) |
                    ((int)blue) |
                    0xff); // hardcode alpha ;
        }
    }
    return temp;
}

我这样用它：

vector<int> NewImage = resizeBilinear(IntData, image_width, image_height, window_width, window_height);

应该返回重新调整大小的图像的RGBA矢量。现在我正在改回RGBA（来自int）

Image.clear();

for(unsigned i=0; i<NewImage.size(); i++)
{
    Image.push_back(NewImage[i] & 255);
    Image.push_back((NewImage[i] >> 8) & 255);
    Image.push_back((NewImage[i] >> 16) & 255);
    Image.push_back(0xff);
}

我得到的是一个黑色的窗口（默认的背景颜色），所以我不知道我错过了什么。如果我注释掉在那里我得到了新的形象，只是转换回RGBA行IntImage，我得到了正确的价值观，所以我不知道这是否是搞砸了就RGBA / INT＆LT;＆GT; INT / RGBA。我现在迷失了。我知道这可以优化/简化，但现在我只想让它发挥作用。

Answer 1

代码中的数组访问不正确：

vector<int> temp(w2 * h2); // initializes the array to contain zeros
...
temp.push_back(...); // appends to the array, leaving the zeros unchanged

你应该覆盖而不是追加;为此，计算数组位置：

temp[i * w2 + j] = ...;

或者，将数组初始化为空状态，并附加你的东西：

vector<int> temp;
temp.reserve(w2 * h2); // reserves some memory; array is still empty
...
temp.push_back(...); // appends to the array

使用C ++和RGB像素矢量调整双线性

1 个答案: