双线性用C和RGBA像素矢量重新调整大小
发布时间:2020-12-16 07:07:32 所属栏目:百科 来源:网络整理
导读:我试图通过使用我发现 here的双线性技术重新调整图像大小,但除了黑色图像我什么也看不到. 所以,首先我用 LodePNG解码我的图像,像素进入矢量 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
我试图通过使用我发现
here的双线性技术重新调整图像大小,但除了黑色图像我什么也看不到.
所以,首先我用 LodePNG解码我的图像,像素进入矢量< 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). 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; } 我这样使用它: 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.如果我注释掉我获得新图像的行并且只是转换回RGBA IntImage我得到正确的值,所以我不知道它是否是混乱的RGBA / int<> INT / RGBA.我现在迷失了.我知道这可以优化/简化,但现在我只想让它工作. 解决方法
您的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,计算阵列位置: 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 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |
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