I am trying enumerate Video Capture format for Logitech camera.I am using this.
I got following entries
MF_MT_FRAME_SIZE 640 x 480
MF_MT_AVG_BITRATE 6619136
MF_MT_COMPRESSED 1
MF_MT_H264_MAX_MB_PER_SEC 245,0,245,0,0,0,0,0,0,0
MF_MT_MAJOR_TYPE MFMediaType_Video
MF_MT_H264_SUPPORTED_USAGES 3
MF_MT_H264_SUPPORTED_RATE_CONTROL_MODES 15
MF_MT_AM_FORMAT_TYPE {2017BE05-6629-4248-AAED-7E1A47BC9B9C}
MF_MT_H264_SUPPORTED_SYNC_FRAME_TYPES 2
MF_MT_MPEG2_LEVEL 40
MF_MT_H264_SIMULCAST_SUPPORT 0
MF_MT_MPEG2_PROFILE 256
MF_MT_FIXED_SIZE_SAMPLES 0
MF_MT_H264_CAPABILITIES 33
MF_MT_FRAME_RATE 30 x 1
MF_MT_PIXEL_ASPECT_RATIO 1 x 1
MF_MT_H264_SUPPORTED_SLICE_MODES 14
MF_MT_ALL_SAMPLES_INDEPENDENT 0
MF_MT_FRAME_RATE_RANGE_MIN 30 x 1
MF_MT_INTERLACE_MODE 2
MF_MT_FRAME_RATE_RANGE_MAX 30 x 1
MF_MT_H264_RESOLUTION_SCALING 3
MF_MT_H264_MAX_CODEC_CONFIG_DELAY 1
MF_MT_SUBTYPE MFVideoFormat_H264_ES
MF_MT_H264_SVC_CAPABILITIES 1
Note: I have modified the function in Media Type Debugging Code as follows.when i run the program i got cElement = 10 and i have put pElemet in for loop to get this value MF_MT_H264_MAX_MB_PER_SEC 245,0,245,0,0,0,0,0,0,0
case VT_VECTOR | VT_UI1:
{
//DBGMSG(L"<<byte array Value>>");
// Item count for the array.
UINT cElement = var.caub.cElems/sizeof(UINT);
// Array pointer.
UINT* pElement = (UINT*)(var.caub.pElems);
for( int i = 0; i < cElement;i++)
DBGMSG(L"%d,", pElement[i]);
}
I am not able to find out what these value signifies
MF_MT_AM_FORMAT_TYPE {2017BE05-6629-4248-AAED-7E1A47BC9B9C}
MF_MT_H264_MAX_MB_PER_SEC 245,0,245,0,0,0,0,0,0,0
MSDN explains value of MF_MT_H264_MAX_MB_PER_SEC attribute:
Data type
UINT32[] stored as UINT8[]
Hence, array of bytes is the expected formatting.
The value of the attribute is an array of UINT32 values, which correspond to the following fields in the UVC 1.5 H.264 video format descriptor.
You have:
dwMaxMBperSecOneResolutionNoScalability
Specifies the maximum macroblock processing rate allowed for
non-scalable Advanced Video Coding (AVC) streams, summing up across
all layers when all layers have the same resolution.
16056565
dwMaxMBperSecTwoResolutionsNoScalability
Specifies the maximum macroblock processing rate allowed for
non-scalable AVC streams, summing up across all layers when all layers
consist of two different resolutions.
0
Media Type GUID "2017be05-6629-4248-aaed-7e1a47bc9b9c" means FORMAT_UVCH264Video
You can then cast the pbFormat struct to KS_H264VIDEOINFO*
Related
I'm working with the TMediaPlayer1 control in an FMX app using C++ Builder 10.2 Version 25.0.29899.2631. The code below runs fine in Win32 and gives the expected result after loading an mp3 file that is 35 minutes, 16 seconds long.
When i run this same code targeting iOS i get the following error:
[bcciosarm64 Error] Unit1.cpp(337): use of overloaded operator '/' is ambiguous (with operand types 'Fmx::Media::TMediaTime' and 'int')
Here is my code that takes the TMediaPlayer1->Duration and converts it to min:sec,
UnicodeString S = System::Ioutils::TPath::Combine(System::Ioutils::TPath::GetDocumentsPath(),"43506.mp3");
if (FileExists(S)) {
MediaPlayer1->FileName = S;
int sec = MediaPlayer1->Duration / 10000000; // <-- this is problem line
int min = sec / 60;
sec = sec - (60 * min);
lblEndTime->Text = IntToStr(min) + ":" + IntToStr(sec);
}
How should i be doing that division?
UPDATE 1: I fumbled around and figured out how to see the values with this code below. When i run on Win32 i get 21169987500 for the Duration (35 min, 16 seconds) and i get 10000000 for MediaTimeScale - both correct. When i run on iOS i get 0 for Duration and 10000000 for MediaTimeScale. But, if i start the audio playing (e.g. MediaPlayer1->Play();) first and THEN run those 2 showmessages i get the correct result for Duration.
MediaPlayer1->FileName = S; // load the mp3
ShowMessage(IntToStr((__int64) Form1->MediaPlayer1->Media->Duration));
ShowMessage(IntToStr((__int64) MediaTimeScale));
It looks like the Duration does not get set on iOS until the audio actually starts playing. I tried a 5 second delay after setting MediaPlayer1->Filename but that doesn't work. I tried a MediaPlayer1->Play(); followed by MediaPlayer->Stop(); but that didn't work either.
Why isn't Duration set when the FileName is assigned? I'd like to show the Duration before the user ever starts playing the audio.
I think that I'm exceeding EPS max line limits:
I'm generating an eps programmatically that consists of a grid of pictures.
My EPS has this structure:
%!PS-Adobe-3.0 EPSF-3.0
.
.
%%BeginProlog
%%EndProlog
%%Page: 1 1
%%Begin Raster Image. Index: 0
save
449 2576 translate
0 rotate
-282 -304 translate
[1 0 0 1 0 0] concat
0 0 translate
[1 0 0 1 0 0] concat
0 0 translate
userdict begin
DisplayImage
0 0
564 608
12
564 608
0
0
FBDBB9FBDCBCFDDBBAFFD8B2FFD7A9FED4A1FCD29CFDD09EFED0A2FFD0A6FFCDA3FFCBA0FFCBA0...
EED79CEBD09CEDD19EEED2A1EFD3A3F0D4A5F0D4A6F0D4A7F1D4A4F3D4A0F3D49F
end
restore
%%End Raster Image
%%Begin Raster Image. Index: 1
.
.
end
restore
%%End Raster Image
%%Begin Raster Image. Index: 2
etc
So the thing is if I write up to 4 images to the EPS, everything works fine, but when I try to write a 5th, the eps won't open on any EPS viewer including Adobe Illustrator (the operation cannot complete because of an unknown error).
I tried using different images to make sure that the particular images were ok and I got the same result, as long as I'm writing 4 images (105825 lines file) everything works. But when I use 5 (132253 lines file) it fails.
Is it possible that I'm exceeding the maximum line limit for EPS?
This are the files in question if you'd like to analyze them:
the one that works - >https://files.fm/u/bfn2d32m and the one that doesn't ->
https://files.fm/u/4gbybr3y
There's no 'line limit' in PostScript or EPS, so you can't be hitting that.
When I run your file through Ghostscript it throws an error /undefined in yImage (I'd suggest you debug PostScript using a proper PostScript interpreter, not Adobe Illustrator).
This suggests to me that one of your images is using more data than you have supplied, so the interpreter runs off the end of the data, consuming parts of the program, until it has read sufficient bytes from currentfile to satisfy the request. At that point is starts processing the file as PostScript again, but the file pointer now points to the 'yImage' of the next 'DisplayImage'. Since you haven't defined a 'yImage' key, naturally this gives you an 'undefined' error.
From your description, this would seem likely to be the 4th image, since adding the 5th throws the error. Note that if your program terminates without supplying enough data (so the interpreter reaches EOF) then the data supplied will be drawn. So it might look like your 4th image is correct even when it isn't, provided it isn't followed by any further program code.
A style note; PostScript is a stack-based language, so normally one would proceed by pushing values onto the stack and reading them from there, instead of executing the 'token' operator.
So your input would be more like :
0 0
564 608
12
564 608
0
0
DisplayImage
FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
...
And the DisplayImage code would be :
/DisplayImage
{
%
% Display a DirectClass or PseudoClass image.
%
% Parameters:
% x & y translation.
% x & y scale.
% label pointsize.
% image label.
% image columns & rows.
% class: 0-DirectClass or 1-PseudoClass.
% compression: 0-none or 1-RunlengthEncoded.
% hex color packets.
%
gsave
/buffer 512 string def
/byte 1 string def
/color_packet 3 string def
/pixels 768 string def
/compression exch def
/class exch def
/rows exch def
/columns exch def
/pointsize exch def
scale
translate
This avoids you having to use token at all for the scale and translate operations for instance.
Very simple issue here I suspect, I am completely new to Matlab (1st day) and have been unable to resolve this by consulting the documentation.
As part of a larger problem, I need to get the dimensions of an image into two variables (row_obj and col_obj for the image object).
According to the documentation:
[m,n] = size(obj)
m: The number of rows in obj.
n: The number of columns in obj.
So, following that, I wrote:
[row_obj, col_obj] = size(object);
disp(row_obj);
disp(col_obj);
disp(size(object));
Which produced the output:
> >> call_i_spy
> 21
>
> 81
>
> 21 27 3
It appears row_obj is correct. If disp(size(object)) produces 21 27 3, then why is col_obj not 27 (the value I need)? What is [row_obj, col_obj] = size(object); actually doing?
I have few mp3 files as binary strings with same number of channels and same sample rate. I need to concatenate them in memory without using command line tools.
Currently I just do string concatenation, like this:
out = ''
mp3s.each { |mp3| out << mp3 }
Audio players can play the result, but with some warnings, because mp3 headers were not handled correctly as far as I understand.
Is there a way to proceed the concatenation in more correct way?
After reading this article about MP3 in russian I came up with solution.
You must be able to get complete ID3 specification at http://id3.org/ but it seems to be down at the moment.
Usually Mp3 file have the next format:
[ID3 head(10 bytes) | ID3 tags | MP3 frames ]
ID3 is not part of MP3 format, but it's kind of container which is used to put information like artists, albums, etc...
The audio data itself are stored in MP3 frames.Every frame starts with 4 bytes header which provides meta info (codecs, bitrate, etc).
Every frame has fixed size. So if there are not enough samples at the end of last frame, coder adds silence to make frame have necessary size. I also found there chunks like
LAME3.97 (name and version of coder).
So, all we need to do is to get rid of ID3 container. The following solution works for me perfect, no warnings anymore and out file became smaller:
# Length of header that describes ID3 container
ID3_HEADER_SIZE = 10
# Get size of ID3 container.
# Length is stored in 4 bytes, and the 7th bit of every byte is ignored.
#
# Example:
# Hex: 00 00 07 76
# Bin: 00000000 00000000 00000111 01110110
# Real bin: 111 1110110
# Real dec: 1014
#
def get_id3_size(header)
result = 0
str = header[6..9]
# Read 4 size bytes from left to right applying bit mask to exclude 7th bit
# in every byte.
4.times do |i|
result += (str[i].ord & 0x7F) * (2 ** (7 * (3-i)))
end
result
end
def strip_mp3!(raw_mp3)
# 10 bytes that describe ID3 container.
id3_header = raw_mp3[0...ID3_HEADER_SIZE]
id3_size = get_id3_size(id3_header)
# Offset from which mp3 frames start
offset = id3_size + ID3_HEADER_SIZE
# Get rid of ID3 container
raw_mp3.slice!(0...offset)
raw_mp3
end
# Read raw mp3s
hi = File.binread('hi.mp3')
bye = File.binread('bye.mp3')
# Get rid of ID3 tags
strip_mp3!(hi)
strip_mp3!(bye)
# Concatenate mp3 frames
hi << bye
# Save result to disk
File.binwrite('out.mp3', hi)
How can I know if a TIFF image is in the format CCITT T.6(Group 4)?
You can use this (C#) code example.
It returns a value indicating the compression type:
1: no compression
2: CCITT Group 3
3: Facsimile-compatible CCITT Group 3
4: CCITT Group 4 (T.6)
5: LZW
public static int GetCompressionType(Image image)
{
int compressionTagIndex = Array.IndexOf(image.PropertyIdList, 0x103);
PropertyItem compressionTag = image.PropertyItems[compressionTagIndex];
return BitConverter.ToInt16(compressionTag.Value, 0);
}
You can check these links
The TIFF File Format
TIFF Tag Compression
TIFF File Format Summary
The tag 259 (hex 0x0103) store the info about the Compression method.
--- Compression
Tag = 259 (103)
Type = word
N = 1
Default = 1.
1 = No compression, but pack data into bytes as tightly as possible, with no
unused bits except at the end of a row. The bytes are stored as an array
of bytes, for BitsPerSample <= 8, word if BitsPerSample > 8 and <= 16, and
dword if BitsPerSample > 16 and <= 32. The byte ordering of data >8 bits
must be consistent with that specified in the TIFF file header (bytes 0
and 1). Rows are required to begin on byte boundaries.
2 = CCITT Group 3 1-Dimensional Modified Huffman run length encoding.
See ALGRTHMS.txt BitsPerSample must be 1, since this type of compression
is defined only for bilevel images (like FAX images...)
3 = Facsimile-compatible CCITT Group 3, exactly as specified in
"Standardization of Group 3 facsimile apparatus for document
transmission," Recommendation T.4, Volume VII, Fascicle VII.3,
Terminal Equipment and Protocols for Telematic Services, The
International Telegraph and Telephone Consultative Committee
(CCITT), Geneva, 1985, pages 16 through 31. Each strip must
begin on a byte boundary. (But recall that an image can be a
single strip.) Rows that are not the first row of a strip are
not required to begin on a byte boundary. The data is stored as
bytes, not words - byte-reversal is not allowed. See the
Group3Options field for Group 3 options such as 1D vs 2D coding.
4 = Facsimile-compatible CCITT Group 4, exactly as specified in
"Facsimile Coding Schemes and Coding Control Functions for Group
4 Facsimile Apparatus," Recommendation T.6, Volume VII, Fascicle
VII.3, Terminal Equipment and Protocols for Telematic Services,
The International Telegraph and Telephone Consultative Committee
(CCITT), Geneva, 1985, pages 40 through 48. Each strip must
begin on a byte boundary. Rows that are not the first row of a
strip are not required to begin on a byte boundary. The data is
stored as bytes, not words. See the Group4Options field for
Group 4 options.
5 = LZW Compression, for grayscale, mapped color, and full color images.
You can run identify -verbose from the ImageMagick suite on the image. Look for "Compression: Group4" in the output.
UPDATE:
SO, I downloaded the libtiff library from the link I mentioned before, and from what I've seen, you can do the following: (untested)
int isTIFF_T6(const char* filename)
{
TIFF* tif= TIFFOpen(filename,"r");
TIFFDirectory *td = &tif->tif_dir;
if(td->td_compression == COMPRESSION_CCITTFAX4) return 1;
return 0;
}
PREVIOUS:
This page has a lot of information about this format and links to some code in C:
Here's an excerpt:
The following paper covers T.4, T.6
and JBIG:
"Review of standards for electronic
imaging for facsimile systems" in
Journal of Electronic Imaging, Vol. 1,
No. 1, pp. 5-21, January 1992.
Source code can be obtained as part of
a TIFF toolkit - TIFF image
compression techniques for binary
images include CCITT T.4 and T.6:
ftp://ftp.sgi.com/graphics/tiff/tiff-v3.4beta035-tar.gz
Contact: sam#engr.sgi.com
Read more: http://www.faqs.org/faqs/compression-faq/part1/section-16.html#ixzz0TYLGKnHI