不管是哪种 Windows DIB 子类型,这种文件格式总是由 14 位文件头和 40 位信息头组成。这两个标头精确包含有关文件的存储内容和存储次序的信息。有关标头中每一项的确切含义,请参考 Microsoft Software Development Kit (SDK)。文件其余部分的内容随信息头中数据的不同而不同。
我们看一下本文要处理的两种子类型。24 位格式很简单:RGB(红-绿-蓝)颜色值(3 个字节,并按 BGR 排序)紧接在信息头之后。但是,每个扫描行都被补足到 4 个字节。按照说明文档(请参阅 Microsoft SDK)的说法,这种“补足”是为了优化 Windows 位图绘图 API。同时,底部的扫描行是文件中的第一项内容 -- 因此相对普通的图形坐标系统(其矢量方向的正向分别为向下和向右)而言,必须从后向前读取图像。
// 解释数据。 int nsize = (((int)bf[5]&0xff)< < 24) (((int)bf[4]&0xff)< < 16) (((int)bf[3]&0xff)< < 8) (int)bf[2]&0xff; System.out.println("File type is :"+(char)bf[0]+(char)bf[1]); System.out.println("Size of file is :"+nsize);
int nbisize = (((int)bi[3]&0xff)< < 24) (((int)bi[2]&0xff)< < 16) (((int)bi[1]&0xff)< < 8) (int)bi[0]&0xff; System.out.println("Size of bitmapinfoheader is :"+nbisize);
int nwidth = (((int)bi[7]&0xff)< < 24) (((int)bi[6]&0xff)< < 16) (((int)bi[5]&0xff)< < 8) (int)bi[4]&0xff; System.out.println("Width is :"+nwidth);
int nheight = (((int)bi[11]&0xff)< < 24) (((int)bi[10]&0xff)< < 16) (((int)bi[9]&0xff)< < 8) (int)bi[8]&0xff; System.out.println("Height is :"+nheight);
int nplanes = (((int)bi[13]&0xff)< < 8) (int)bi[12]&0xff; System.out.println("Planes is :"+nplanes);
int nbitcount = (((int)bi[15]&0xff)< < 8) (int)bi[14]&0xff; System.out.println("BitCount is :"+nbitcount);
// 查找表明压缩的非零值 int ncompression = (((int)bi[19])< < 24) (((int)bi[18])< < 16) (((int)bi[17])< < 8) (int)bi[16]; System.out.println("Compression is :"+ncompression);
int nsizeimage = (((int)bi[23]&0xff)< < 24) (((int)bi[22]&0xff)< < 16) (((int)bi[21]&0xff)< < 8) (int)bi[20]&0xff; System.out.println("SizeImage is :"+nsizeimage);
int nXPm = (((int)bi[27]&0xff)< < 24) (((int)bi[26]&0xff)< < 16) (((int)bi[25]&0xff)< < 8) (int)bi[24]&0xff; System.out.println("X-Pixels per meter is :"+nxpm);
int nypm = (((int)bi[31]&0xff)< < 24) (((int)bi[30]&0xff)< < 16) (((int)bi[29]&0xff)< < 8) (int)bi[28]&0xff; System.out.println("Y-Pixels per meter is :"+nypm);
int nclrused = (((int)bi[35]&0xff)< < 24) (((int)bi[34]&0xff)< < 16) (((int)bi[33]&0xff)< < 8) (int)bi[32]&0xff; System.out.println("Colors used are :"+nclrused);
int nclrimp = (((int)bi[39]&0xff)< < 24) (((int)bi[38]&0xff)< < 16) (((int)bi[37]&0xff)< < 8) (int)bi[36]&0xff; System.out.println("Colors important are :"+nclrimp);
if (nbitcount==24) { // 24 位格式不包含调色板数据,但扫描行被补足到 // 4 个字节。 int npad = (nsizeimage / nheight) - nwidth * 3; int ndata[] = new int [nheight * nwidth]; byte brgb[] = new byte [( nwidth + npad) * 3 * nheight]; fs.read (brgb, 0, (nwidth + npad) * 3 * nheight); int nindex = 0; for (int j = 0; j < nheight; j++) { for (int i = 0; i < nwidth; i++) { ndata [nwidth * (nheight - j - 1) + i] = (255&0xff)< < 24 (((int)brgb[nindex+2]&0xff)< < 16) (((int)brgb[nindex+1]&0xff)< < 8) (int)brgb[nindex]&0xff; // System.out.println("Encoded Color at (" +i+","+j+")is:"+nrgb+" (R,G,B)= (" +((int)(brgb[2]) & 0xff)+"," +((int)brgb[1]&0xff)+"," +((int)brgb[0]&0xff)+")"); nindex += 3; } nindex += npad; }
image = createImage ( new MemoryImageSource (nwidth, nheight, ndata, 0, nwidth)); } else if (nbitcount == 8) { // 必须确定颜色数。假如 clrsused 参数大于 0, // 则颜色数由它决定。假如它等于 0,则根据 // bitsperpixel 计算颜色数。 int nNumColors = 0; if (nclrused > 0) { nNumColors = nclrused; } else { nNumColors = (1&0xff)< < nbitcount; } System.out.println("The number of Colors is"+nNumColors);