用C 和RGBA像素的向量重新尺寸

Question

我正在尝试使用我在这里找到的双线性技术来重新尺寸大小，但除了黑色图像之外，我什么也看不到。因此，首先，我将图像用lodepng解码，然后将像素分为vector<unsigned char>变量。它说它们被存储为rgbargba，但是当我尝试将图像应用于X11窗口时，我意识到它们被存储为Bgrabgra。我不知道是否会更改顺序或lodepng解码器的X11 API。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;
}

因此，现在我试图在将图像大小应用于窗口之前更改图像的大小。大小将是窗户的大小（拉伸）。我首先尝试将RGBA转换为INT值，例如：

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此功能，该链接应该进行插值：

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;
}

我这样使用它：

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

应该将我的RGBA矢量归还重新尺寸的图像的RGBA矢量。现在，我正在换回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窗口（默认背景颜色），所以我不知道我缺少什么。如果我要评论获得新映像的行，然后将其转换回RGBA，则可以将IntImage转换回RGBA，我会得到正确的值，因此我不知道这是否是混乱的RGBA/INT＆lt;> int/rgba。我现在就迷路了。我知道这可以优化/简化，但现在我只想使它起作用。

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

您应该覆盖而不是附加；为此，计算数组位置：

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

或者，将数组初始化为空状态，然后附加您的东西：

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