Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
583 views
in Technique[技术] by (71.8m points)

c# - What quality level does Image.Save() use for jpeg files?

I just got a real surprise when I loaded a jpg file and turned around and saved it with a quality of 100 and the size was almost 4x the original. To further investigate I open and saved without explicitly setting the quality and the file size was exactly the same. I figured this was because nothing changed so it's just writing the exact same bits back to a file. To test this assumption I drew a big fat line diagonally across the image and saved again without setting quality (this time I expected the file to jump up because it would be "dirty") but it decreased ~10Kb!

At this point I really don't understand what is happening when I simply call Image.Save() w/out specifying a compression quality. How is the file size so close (after the image is modified) to the original size when no quality is set yet when I set quality to 100 (basically no compression) the file size is several times larger than the original?

I've read the documentation on Image.Save() and it's lacking any detail about what is happening behind the scenes. I've googled every which way I can think of but I can't find any additional information that would explain what I'm seeing. I have been working for 31 hours straight so maybe I'm missing something obvious ;0)

All of this has come about while I implement some library methods to save images to a database. I've overloaded our "SaveImage" method to allow explicitly setting a quality and during my testing I came across the odd (to me) results explained above. Any light you can shed will be appreciated.

Here is some code that will illustrate what I'm experiencing:

string filename = @"C:empimage testinghh.jpg";
string destPath = @"C:empimage testing";

using(Image image = Image.FromFile(filename))
{
    ImageCodecInfo codecInfo = ImageUtils.GetEncoderInfo(ImageFormat.Jpeg);

    //  Set the quality
    EncoderParameters parameters = new EncoderParameters(1);

    // Quality: 10
    parameters.Param[0] = new EncoderParameter(
        System.Drawing.Imaging.Encoder.Quality, 10L);
    image.Save(destPath + "10.jpg", codecInfo, parameters);

    // Quality: 75
    parameters.Param[0] = new EncoderParameter(
        System.Drawing.Imaging.Encoder.Quality, 75L);
    image.Save(destPath + "75.jpg", codecInfo, parameters);

    // Quality: 100
    parameters.Param[0] = new EncoderParameter(
        System.Drawing.Imaging.Encoder.Quality, 100L);
    image.Save(destPath + "100.jpg", codecInfo, parameters);

    //  default
    image.Save(destPath + "default.jpg", ImageFormat.Jpeg);

    //  Big line across image
    using (Graphics g = Graphics.FromImage(image))
    {
        using(Pen pen = new Pen(Color.Red, 50F))
        {
            g.DrawLine(pen, 0, 0, image.Width, image.Height);
        }
    }

    image.Save(destPath + "big red line.jpg", ImageFormat.Jpeg);
}

public static ImageCodecInfo GetEncoderInfo(ImageFormat format)
{
    return ImageCodecInfo.GetImageEncoders().ToList().Find(delegate(ImageCodecInfo codec)
    {
        return codec.FormatID == format.Guid;
    });
}
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

Using reflector, it turns out Image.Save() boils down to the GDI+ function GdipSaveImageToFile, with the encoderParams NULL. So I think the question is what the JPEG encoder does when it gets a null encoderParams. 75% has been suggested here, but I can't find any solid reference.

EDIT You could probably find out for yourself by running your program above for quality values of 1..100 and comparing them with the jpg saved with the default quality (using, say, fc.exe /B)


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...