libjpeg:如何获得正确的图像解压缩大小?

时间:2012-10-18 14:53:27

标签: c++ jpeg compression libjpeg

我有这样一个通常的代码:

struct jpeg_decompress_struct cinfo;
jpeg_create_decompress(&cinfo);
jpeg_stdio_src(&cinfo, infile);
jpeg_read_header(&cinfo, TRUE);

cinfo.scale_num = ?;
cinfo.scale_denom = ?;

needed_width = cinfo.output_width;
needed_height = cinfo.output_height;

如何选取 scale_num scale_denum 参数以将图像缩放到所需的尺寸。我希望例如将它缩小一倍。

如果我设置scale_num = 1scale_denom = 2。结果是:(1802 x 1237)至(258 x 194)

文档说:

Scale the image by the fraction scale_num/scale_denom.  Default is
1/1, or no scaling.  Currently, the only supported scaling ratios
are 1/1, 1/2, 1/4, and 1/8.

但是当我设定这样的比例时,我得不到所需的结果。

所以问题是:如何设置scale_numscale_denom以获得与所需尺寸最相似的图片。

1 个答案:

答案 0 :(得分:4)

您应该调用jpeg_calc_output_dimensions(cinfo)来获取正确的output_width和output_height值。

计算系数scale_denom我通常做这样的事情:

unsigned int intlog2(unsigned int val) {
 int targetlevel = 0;
 while (val >>= 1) ++targetlevel;
 return targetlevel;
}

// ....
// for example, 
const int width = 5184;
const int height = 3456;

unsigned int needed_width = 640;
unsigned int needed_height = 480;

unsigned int wdeg = intlog2(width / needed_width);
unsigned int hdeg = intlog2(height / needed_height);

unsigned int scale_den = std::min(1 << (std::min)(wdeg, hdeg), 8);

unsigned int result_width = width / scale_den;
unsigned int result_height = height / scale_den;