Analyze peculiar avcC atom structure

I need some help to understand the avcC atom structure of a particular mp4 sample I am trying to analyze.

Hex dump:

00 00 00 38 61 76 63 43 01 64 00 1F FF E1 00 1C 67 64 00 1F AC D9 80 50 05 BB 01 6A 02 02 02 80 00 00 03 00 80 00 00 1E 07 8C 18 CD 01 00 05 68 E9 7B 2C 8B FD F8 F8 00 00 00 00 13 63 6F 6C 72

This is what I understand from the above:

00 00 00 38 Size of avcC atom

61 76 63 43 avcC signature

01 configurationVersion

64 AVCProfileIndication

00 profile_compatibility

1F AVCLevelIndication

FF 111111b + lengthSizeMinusOne

E1 111b + numOfSequenceParameterSets (in this case, 1 SPS)

00 1C SPS length (in this case, 28 bytes)

67 64 00 1F AC D9 80 50 05 BB 01 6A 02 02 02 80 00 00 03 00 80 00 00 1E 07 8C 18 CD SPS data (28 bytes as per above)

01 numOfPictureParameterSets (in this case, 1 PPS)

00 05 PPS length

This is where the problem begins. Based on the PPS length given by the previous bytes, the next 5 bytes should be the PPS data: 68 E9 7B 2C 8B

However according to the avcC header, the total length of the atom is 56 bytes (0x38), which means that the following 4 bytes should be included: FD F8 F8 00

But the problem is that the PPS length is given as 5 bytes (0x05). So what exactly are these final 4 bytes?

Then follows the header of the colr atom:

00 00 00 13 size of colr atom

63 6F 6C 72 colr signature

Which I have checked and is indeed 19 bytes in length (0x13).

The problem is with the avcC atom and with that particular mp4 sample I am analyzing (I've checked other samples too and they didn't have this peculiarity).

You can find the sample here.

EDIT

mp4info tool from the bento4 suite reports the following as the avcC atom's size: 8+48

And mp4dump reports:

AVC SPS: [6764001facd9805005bb016a02020280000003008000001e078c18cd]

AVC PPS: [68e97b2c8b]

So it correctly reports the total size of the atom as 56 bytes (0x38) based on what is found in the avcC header, but the SPS/PPS data are analyzed the same way as above. I still don't understand what the final 4 bytes are or where do they belong.

Solution

I dind't get any answer but fortunately a bit more careful reading of ISO 14496-15 solved this issue:

if( profile_idc == 100 || profile_idc == 110 ||
    profile_idc == 122 || profile_idc == 144 )
{
   bit(6) reserved = ‘111111’b;
   unsigned int(2) chroma_format;
   bit(5) reserved = ‘11111’b;
   unsigned int(3) bit_depth_luma_minus8;
   bit(5) reserved = ‘11111’b;
   unsigned int(3) bit_depth_chroma_minus8;
   unsigned int(8) numOfSequenceParameterSetExt;
   for (i=0; i< numOfSequenceParameterSetExt; i++) {
      unsigned int(16) sequenceParameterSetExtLength;
      bit(8*sequenceParameterSetExtLength) sequenceParameterSetExtNALUnit;
   }
}

Apparently a sequence of 4+ bytes may exist at the end of an avcC atom depending on the profile used. In my sample above the profile is 100 (0x64), hence it meets the criteria. So the last 4 bytes are:

FD = bits 111111 are reserved, remaining 01 means chroma subsampling 4:2:0

F8 = bits 11111 are reserved, remaining 000 means luma bit depth is 8

F8 = bits 11111 are reserved, remaining 000 means chroma bit depth is 8

00 = zero SPS extensions