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So I'm working on a WinRT app in VB.NET that takes images and makes a video (.wmv) out of them.
Since this is not achievable from .NET alone, I'm using a C++ library from here: blogs.msdn.com[^]
I finally got a video file outputted but was disturbed to note a discoloring of the frames.
(See pictures)
Original Image[^]
Video Frame[^]

I believe the problem is with the library but since I'm not at all familiar with C++, I can't figure out what's going on.

Here's my .NET code:
VB
Dim VidGen As New VideoTools.VideoGenerator(640, 480, stream, 33)
Dim imgfile As StorageFile
Dim imgstream As IRandomAccessStream = Await imgfile.OpenAsync(FileAccessMode.Read)
Dim img As New WriteableBitmap(640, 480)
img.SetSource(imgstream)
ImgTest.Source = img
Try
  Dim arr As Byte() = img.ToByteArray()
  VidGen.AppendNewFrame(arr)
...

Here's the C++ code that I'm most directly calling:
C++
void VideoGenerator::AppendNewFrame(const Platform::Array<byte> ^videoFrameBuffer)
{
	auto length = videoFrameBuffer->Length / sizeof(DWORD);
	DWORD *buffer = (DWORD *)(videoFrameBuffer->Data);
	std::unique_ptr<DWORD[]> target(new DWORD[length]);

	for (UINT32 index = 0; index < length; index++)
	{
		DWORD color = buffer[index];
		BYTE b = (BYTE)((color & 0x00FF0000) >> 16);
		BYTE g = (BYTE)((color & 0x0000FF00) >> 8);
		BYTE r = (BYTE)((color & 0x000000FF));

#if ARM
		auto row = index / videoWidth;
		auto targetRow = videoHeight - row - 1;
		auto column = index - (row * videoWidth);
		target[(targetRow * videoWidth) + column] = (r << 16) + (g << 8) + b;
#else
		target[index] = (r << 16) + (g << 8) + b;
#endif
	}

	// Send frame to the sink writer.
	HRESULT hr = WriteFrame(target.get(), rtStart, rtDuration);
	if (FAILED(hr))
	{
		throw Platform::Exception::CreateException(hr);
	}
	rtStart += rtDuration;
}

Any ideas on what's causing the discoloration and how to fix it? Thanks.
Posted

I finally found an alternative to using WriteableBitmap to export to a byte array.
I started using this function to do that:
VB
Public Async Function ImageFileToByteArray(file As StorageFile) As Task(Of [Byte]())
        Dim stream As IRandomAccessStream = Await file.OpenAsync(Windows.Storage.FileAccessMode.Read)
        Dim decoder As BitmapDecoder = Await BitmapDecoder.CreateAsync(stream)
        Dim pixelData As PixelDataProvider = Await decoder.GetPixelDataAsync()
        Return pixelData.DetachPixelData()
End Function

This proved to be highly enlightening as I was able to look at the pixel format of the decoder. It told me that the file I was reading was in BGRA8 format. Armed with this knowledge, I rewrote my blue-red swapper and, voila: success!

So my final code was:
VB
Dim arr As Byte() = Await ImageFileToByteArray(imgfile)
Dim newarr(arr.Length - 1) As Byte
For i As Integer = 0 To arr.Length - 1 Step 4
   'i = B
   'i + 1 = G
   'i + 2 = R
   'i + 3 = A
   newarr(i) = arr(i + 2)
   newarr(i + 1) = arr(i + 1)
   newarr(i + 2) = arr(i)
   newarr(i + 3) = arr(i + 3)
Next
VidGen.AppendNewFrame(newarr)

Thank you, Bill for your help! You set me on the right track!
 
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It appears that red and blue are swapped.

Some image formats, such as the BMP format, store 8-bit colors in BGR order.
Other formats are RGB.

Try swapping the red and blue colors.
 
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Comments
Casey Sheridan 4-Nov-13 20:02pm    
I had previously held such a theory but have not yet gotten the right effect.
More background: my pictures are being made into ARGB byte arrays.

Here's what it looks like with red and blue swapped and leaving A and G in place:
https://skydrive.live.com/redir?resid=6A7E013B8B8FE514!2443&authkey=!AEJszMtaRMLYUQM&v=3
Bill_Hallahan 4-Nov-13 22:04pm    
The code you have posted has red and blue swapped already, so I should have written they need to be unswapped to be clearer! Since there's a lot of blue in the original image, with the colors swapped, there should be a considerable red tinge to the image, so if you swapped the colors correctly, there's more going on. I suspect the video doesn't use 8 bits per color component. It could be either 555 or 565 color format. For the second format, that's 6 bits for green and 5 bits for the red and blue, in either rgb or bgr order.

However, i'm just guessing.

Also, in your last message, I have no idea what you mean by A and G in place. There is no A color component.

Please post the code with the red and blue colors swapped.
Bill_Hallahan 4-Nov-13 22:05pm    
Perhaps by ARBG, you meant 32-bits, with an Alpha channel?
Casey Sheridan 4-Nov-13 22:10pm    
Yes, that's exactly what I meant.

I have, however, figured it out now. I detailed what I did in the solution I posted.
I do want to thank you, though, for pointing me in the right direction. You showed me that I wasn't really off my rocker like I thought I was :)
Bill_Hallahan 5-Nov-13 18:48pm    
Great. Of course, I had no idea what the solution was!

I'm glad me being a sounding board helped! :)

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