No image displayed when setting Image.Source to valid data - wpf

Question

I'm receiving byte[] image-data (from a webcam source, transmitted over network) and want to display it in my WPF Image control.
If I set the resolution to 160x120, I see a nice tiny image being displayed.
For any other resolution though I see nada, nothing, zilch, squat.
If I write the raw bytes to disk I can see that the data is indeed a valid jpg-image.
Just to make sure it's not jpeg-incompatibility I've tested it with png- and bmp-encoding/decoding as well with the same results; still no image.
Anyone got a bright idea ?
private bool OnImage(byte[] inJpg)
{
this.Dispatch(() =>
{
//File.WriteAllBytes("h:\\tmp\\test" + sImageNum++ + ".jpg", inJpg);
using (MemoryStream ms = new MemoryStream(inJpg))
{
BitmapDecoder decoder = new JpegBitmapDecoder(ms, BitmapCreateOptions.None, BitmapCacheOption.Default);
ImageRemoteVideo.Source = decoder.Frames[0];
}
}
);
}

Answer 1

Assuming ImageRemoteVideo is an instance of a WPF Image control, you may be getting bit by some lazy resolve issues. The Image control has "BeginInit" and "EndInit" methods that you should probably be using. Put the BeginInit method before you set the stream, the other after you set it. Also, if I remember right, the Image class supports a StreamSource. You could try setting your MemoryStream (without disposing of it and without using the JpegBitmapDecoder) to that Property. If that works, then see if you can dispose it after the EndInit call.

Answer 2

I have just found a way to make it work, but I'd love it if someone can explain WHY it works :)
First off, I'm using the raw bytestream to create a BitmapImage.
BUT - if I use a "using" clause around the MemoryStream it doesn't work, if I don't it works.
So this doesn't work :
using (MemoryStream byteStream = new MemoryStream(inJpg))
{
BitmapImage image = new BitmapImage();
image.BeginInit();
image.StreamSource = byteStream;
image.EndInit();
ImageRemoteVideo.Source = image;
}
But this does :
MemoryStream byteStream = new MemoryStream(inJpg);
BitmapImage image = new BitmapImage();
image.BeginInit();
image.StreamSource = byteStream;
image.EndInit();
ImageRemoteVideo.Source = image;

Answer 3

Found it. It's like Brannon said a case of delayed loading.
Specifying "CacheOption = BitmapCacheOption.OnLoad;" fixes it !
Cheers !
Working version :
MemoryStream byteStream = new MemoryStream(inJpg);
{
BitmapImage image = new BitmapImage();
image.BeginInit();
image.StreamSource = byteStream;
image.CacheOption = BitmapCacheOption.OnLoad;
image.EndInit();
ImageRemoteVideo.BeginInit();
ImageRemoteVideo.Source = image;
ImageRemoteVideo.EndInit();
}