I have Googled a lot and failed to find the decoding algorithm for the PDF417 barcode used by United States Postal Service. I want to fetch recipient and sender information with tracking number out of it.
I have successfully decoded the FedEx barcode with ANSI MH10.8.3 standard.
My question is, am I scanning the correct barcode (I am not from USA, so I don't know which barcode label USPS applies to their parcels) ? If no, then what barcode should I look for to fetch required information ? If Yes, then how can I decode this barcode ?
Please help,
Thanks.
Edit:
Here is another similar barcode
You should check this webpage:
https://en.wikibooks.org/wiki/International_Postage_Meter_Stamp_Catalog/United_States_of_America,_Part_3
As well as this page:
http://www.stamps.com/postage-online/how-it-works/
Your first barcode seems to have been generated by Endicia software (ID# starts with 071V), while the second example barcode was generated by stamps.com (as printed, and also ID# starts with 062S).
It seems that stamps.com service allows you to print stamps without providing the recipient address. For this reason, the barcode probably doesn't include any information about the recipient.
For the record, the decoded bars are as follows:
0000 50 01 dc 78 0c 00 30 37 31 56 57 6c 08 00 7a 86 | P~~x~~071VWl~~z~ |
0010 09 c5 4e d8 27 00 8a b7 32 01 24 4f 00 00 67 49 | ~~N~'~~~2~$O~~gI |
0020 6d 15 00 b5 c3 00 00 00 00 06 c1 31 02 b9 02 90 | m~~~~~~~~~~1~~~~ |
0030 d0 a4 4a 1c 02 2a 42 8f a7 3f 6d c7 03 ea e5 d7 | ~~J~~*B~~?m~~~~~ |
0040 3c 69 86 3c 50 29 28 32 11 74 6a 7f b4 af c7 90 | <i~<P)(2~tj~~~~~ |
0050 16 c3 90 bb fb 2a fa 4e 78 95 e6 20 69 c7 75 01 | ~~~~~*~Nx~~ i~u~ |
0060 00 00 | ~~ |
and:
0000 05 01 ff ff 00 00 30 36 32 53 3b 47 70 00 f2 ed | ~~~~~~062S;Gp~~~ |
0010 10 00 00 14 1e 00 56 52 33 01 59 33 01 00 00 00 | ~~~~~~VR3~Y3~~~~ |
0020 00 00 00 00 04 00 02 00 00 5c da 00 00 38 30 33 | ~~~~~~~~~\~~~803 |
0030 34 ae 69 57 0d 59 42 1c d4 0b 00 f2 d3 7f 4f f8 | 4~iW~YB~~~~~~~O~ |
0040 ef 69 53 a0 aa fb 9b cf 30 16 13 c3 08 3e 86 4a | ~iS~~~~~0~~~~>~J |
0050 7a e8 4c fe 1f eb 4d 2c 52 05 00 6f 33 01 00 | z~L~~~M,R~~o3~~ |
Bytes 06-09 (0-indexed) is the ID prefix in ASCII.
Bytes 0A-0D is the rest of the ID, encoded in binary in little endian. 3B 47 70 00 is 0x0070473B = 7358267, for the second stamp.
For the second stamp, bytes 5B-5D (6F 33 01) is actually 01 33 6F = 78703, the zip it was posted from. Unfortunately, it doesn't work with the first stamp.
Related
I'm developing a minimal x86-64 OS from scratch and I am attempting to detect memory to be able to map the higher half of the virtual address space to all physical memory available.
From this link: https://www.kernel.org/doc/html/latest/x86/x86_64/mm.html, I think this is what the Linux kernel does also. Probably to be able to reach all physical addresses if the need arises at some point.
ffff888000000000 | -119.5 TB | ffffc87fffffffff | 64 TB | direct mapping of all physical memory (page_offset_base)
I want to do the same in my kernel but I need to detect the amount of physical memory installed currently on my system. I can always use the Memory Map returned by UEFI but this doesn't necessarily tell me how much memory is actually installed.
I'm emulating on QEMU and I thought of locating the SMBIOS table to do that. If I print memory from 0xf0000 to 0xfffff, I don't find the signature of the SMBIOS table:
(gdb) dump memory result.bin 0xf0000 0xfffff
user#user-System-Product-Name:~$ hexdump -C result.bin
00000000 ff ff ff ff ff ff ff ff ff ff ff ff ff ff ff ff |................|
*
0000fd00 ff ff ff ff ff ff ff ff 2e 06 a0 1b 79 c7 82 45 |............y..E|
0000fd10 85 66 33 6a e8 f7 8f 09 08 aa 01 08 f8 02 00 f8 |.f3j............|
0000fd20 0c 00 00 19 00 00 00 00 00 00 00 00 d4 02 00 19 |................|
0000fd30 31 c0 2d 00 10 00 00 3d 00 00 00 ff 72 33 81 78 |1.-....=....r3.x|
0000fd40 10 78 e5 8c 8c 75 eb 81 78 14 3d 8a 1c 4f 75 e2 |.x...u..x.=..Ou.|
0000fd50 81 78 18 99 35 89 61 75 d9 81 78 1c 85 c3 2d d3 |.x..5.au..x...-.|
0000fd60 75 d0 83 78 24 00 75 ca 89 c3 03 58 20 75 c3 eb |u..x$.u....X u..|
0000fd70 09 b8 bf bf bf bf 89 c5 eb fe 89 c5 e9 37 02 00 |.............7..|
0000fd80 00 31 db 89 de 89 e8 66 8b 5d 30 01 d8 72 3b eb |.1.....f.]0..r;.|
0000fd90 03 40 72 36 85 c0 74 32 83 c0 07 72 2d 24 f8 8a |.#r6..t2...r-$..|
0000fda0 58 17 f6 c3 20 74 ea 8b 48 14 81 e1 ff ff ff 00 |X... t..H.......|
0000fdb0 09 c9 74 dd 01 c1 74 02 72 d7 80 78 12 03 75 06 |..t...t.r..x..u.|
0000fdc0 eb 17 85 c0 75 06 89 c8 eb ca 31 c0 89 c6 85 f6 |....u.....1.....|
0000fdd0 75 02 74 fe e9 e4 01 00 00 85 c0 74 5f 83 c0 18 |u.t........t_...|
0000fde0 39 c8 73 58 80 78 03 10 74 1b 80 78 03 12 74 32 |9.sX.x..t..x..t2|
0000fdf0 8b 18 81 e3 ff ff ff 00 01 d8 72 40 83 c0 03 72 |..........r#...r|
0000fe00 3b 24 fc eb db 83 c0 04 66 81 38 4d 5a 75 2d 0f |;$......f.8MZu-.|
0000fe10 b7 58 3c 01 c3 81 3b 50 45 00 00 75 1f 03 43 28 |.X<...;PE..u..C(|
0000fe20 eb 1f 83 c0 04 89 c3 66 81 3b 56 5a 75 0e 03 43 |.......f.;VZu..C|
0000fe30 08 83 c0 28 0f b7 5b 06 29 d8 eb 05 b8 00 00 00 |...(..[.).......|
0000fe40 00 e9 7c ff ff ff eb 60 0f 20 e0 0f ba e8 05 0f |..|....`. ......|
0000fe50 22 e0 b9 80 00 00 c0 0f 32 0f ba e8 08 0f 30 0f |".......2.....0.|
0000fe60 20 c0 0f ba e8 1f 0f 22 c0 ea 70 fe ff ff 18 00 | ......"..p.....|
0000fe70 e9 4d 01 00 00 b8 00 00 00 80 0f a2 3d 1f 00 00 |.M..........=...|
0000fe80 80 7c 21 b8 1f 00 00 80 0f a2 0f ba e0 01 73 14 |.|!...........s.|
0000fe90 b9 31 01 01 c0 0f 32 0f ba e0 00 73 07 89 d8 83 |.1....2....s....|
0000fea0 e0 3f eb 02 31 c0 eb 02 eb cb 31 d2 85 c0 74 06 |.?..1.....1...t.|
0000feb0 83 e8 20 0f ab c2 b9 00 18 00 00 31 c0 89 04 8d |.. ........1....|
0000fec0 fc ff 7f 00 e2 f7 c7 05 00 00 80 00 23 10 80 00 |............#...|
0000fed0 89 15 04 00 80 00 c7 05 00 10 80 00 23 20 80 00 |............# ..|
0000fee0 89 15 04 10 80 00 c7 05 08 10 80 00 23 30 80 00 |............#0..|
0000fef0 89 15 0c 10 80 00 c7 05 10 10 80 00 23 40 80 00 |............##..|
0000ff00 89 15 14 10 80 00 c7 05 18 10 80 00 23 50 80 00 |............#P..|
0000ff10 89 15 1c 10 80 00 b9 00 08 00 00 89 c8 48 c1 e0 |.............H..|
0000ff20 15 05 e3 00 00 00 89 04 cd f8 1f 80 00 89 14 cd |................|
0000ff30 fc 1f 80 00 e2 e5 b8 00 00 80 00 0f 22 d8 e9 05 |............"...|
0000ff40 ff ff ff fa bb 00 f0 8e db bb 7a ff 2e 66 0f 01 |..........z..f..|
0000ff50 17 66 b8 23 00 00 40 0f 22 c0 66 ea 62 ff ff ff |.f.#..#.".f.b...|
0000ff60 10 00 b8 40 06 00 00 0f 22 e0 66 b8 08 00 8e d8 |...#....".f.....|
0000ff70 8e c0 8e e0 8e e8 8e d0 eb 39 1f 00 80 ff ff ff |.........9......|
0000ff80 00 00 00 00 00 00 00 00 ff ff 00 00 00 93 cf 00 |................|
0000ff90 ff ff 00 00 00 9b cf 00 ff ff 00 00 00 9b af 00 |................|
0000ffa0 bf 42 50 eb 0a bf 41 50 eb 05 66 89 c4 eb 02 eb |.BP...AP..f.....|
0000ffb0 f9 eb 90 e9 78 fd ff ff e9 c4 fd ff ff e9 84 fe |....x...........|
0000ffc0 ff ff b8 ff ff ff ff 48 21 c6 48 21 c5 48 21 c4 |.......H!.H!.H!.|
0000ffd0 48 89 e0 ff e6 90 90 90 90 90 90 90 90 90 90 90 |H...............|
0000ffe0 eb c3 90 90 90 90 90 90 00 00 00 00 56 54 46 00 |............VTF.|
0000fff0 90 90 eb ac 90 90 90 90 90 90 90 90 90 90 90 |...............|
0000ffff
I did try to add the -smbios type=0 flag when I launch QEMU from the command line.
I was wondering how the Linux kernel, when it runs within QEMU, does to detect memory and hardware. Does it use ACPI tables instead? I think SMBIOS is much more easy to use.
Is SMBIOS reliable enough so that operating-systems that run on newer hardware can assume its presence?
I'm currently working on old system that uses RDP. According to 4.1.4 Server MCS Connect Response PDU with GCC Conference Create Response described in [MS-RDPBCGR], packet is containing modulus, which should be part of RSA key. And I need to know where this came from because I need to decrypt some RDP packets stored as log.
First thing I've done is looking up certificates by using mmc. But there was no certificate matching with modulus. Even if I issued new self-signed certificate, there was no luck. Modulus is not changing by it.
More specifically, this is response packet from testing server(VM) containing modulus.
0000: 03 00 02 15 02 f0 80 7f 66 82 02 09 0a 01 00 02 | ......f......
0016: 01 00 30 1a 02 01 22 02 01 03 02 01 00 02 01 01 | ..0...".........
0032: 02 01 00 02 01 01 02 03 00 ff f8 02 01 02 04 82 | .............
0048: 01 e3 00 05 00 14 7c 00 01 2a 14 76 0a 01 01 00 | .....|..*.v....
0064: 01 c0 00 4d 63 44 6e 81 cc 01 0c 10 00 0c 00 08 | ..McDn.......
0080: 00 00 00 00 00 04 00 00 00 03 0c 10 00 eb 03 04 | ...............
0096: 00 ec 03 ed 03 ee 03 ef 03 02 0c ac 01 02 00 00 | ...........
0112: 00 02 00 00 00 20 00 00 00 78 01 00 00 bb e4 de | ..... ...x...
0128: 58 1a 05 8f 26 89 f8 94 0b 88 d4 79 d4 00 ac bf | X..&.y.
0144: e0 07 72 3a e5 9b 17 7f 17 d6 18 92 7f 01 00 00 | .r:........
0160: 00 01 00 00 00 01 00 00 00 06 00 1c 01 52 53 41 | .............RSA
0176: 31 08 01 00 00 00 08 00 00 ff 00 00 00 01 00 01 | 1..............
0192: 00 2d 13 bc 1d a9 5b c8 60 9b be 66 61 ab 09 13 | .-..[`fa..
0208: 4e 0a 1f 64 27 72 df 92 18 42 ea 2c 05 5d 0d a7 | N..d'r..B,.].
0224: f7 06 51 5d 22 2e 4a fa 03 c5 8d 52 47 7c fa 13 | .Q]".J..RG|.
0240: ec dd bb 81 15 50 4b b3 f0 7b e4 75 0e e6 0d b5 | ..PK{u..
0256: ab d2 4a 9c ab f6 8c 83 a3 53 0b 87 b1 07 fc 0f | JS...
0272: 29 12 f4 c8 18 fb 9f 6d 29 10 34 af 34 d0 ca 8d | )..m).44.
0288: 48 a9 2e 9e 85 9a 39 d6 6c be cb f3 36 75 60 a5 | H.9l6u`
0304: 56 a5 a3 f5 b0 6f af c3 8e 5b 03 11 e4 27 27 bf | Vo.[..''
0320: a0 05 51 aa f1 8d 84 11 53 43 59 b8 83 4f f2 2d | .Q.SCYO-
0336: 40 44 b1 f9 5a 5b e6 2d 32 e4 d8 ef 2a 5a f8 01 | #DZ[-2*Z.
0352: 08 7a 68 a0 05 e2 5b fe 50 b5 38 cd a6 f0 ef e0 | .zh.[P8.
0368: c4 6f 4e f3 f1 9d 0a 89 ce 79 4e 3d 6f e3 a2 b3 | oN.yN=o.
0384: c7 fd dc b2 d8 c6 76 e8 79 67 ca fe 71 5d a5 3d | .vygq]=
0400: d3 40 c4 a4 28 5c 11 b7 2a 51 cd 65 e4 5f fc 2a | #.(\.*Qe_*
0416: bf 4c b1 e0 96 89 05 4b c6 72 1a 62 eb a2 51 0d | L.Kr.bQ.
0432: 45 2f 23 27 67 0e a8 c6 12 ed 81 ee 09 58 10 02 | E/#'g...X..
0448: b2 00 00 00 00 00 00 00 00 08 00 48 00 e9 95 02 | ..........H..
0464: 48 e7 84 d6 fc 60 cd 29 b2 91 7c f4 e8 b4 36 5d | H`)|6]
0480: e5 5e b4 90 d4 d4 5d 6a a1 42 69 c6 4e 5c 87 f2 | ^]jBiN\
0496: 0a cd 86 f5 64 e3 4d 61 60 0a 17 c2 f8 94 93 83 | ..dMa`..
0512: cf 23 7d c4 a3 07 ad f0 b6 bc 1a b1 00 00 00 00 | #}.......
0528: 00 00 00 00 00 | .....
Public exponent is 01 00 01 00, modulus is 2d 13 bc 1d ... 58 10 02 b2 with additional 8 bytes of zero-padding.
After that, if I know what private exponent is, then I can decrypt Client Random and generate session key.
But as I've mentioned, I can't find where modulus is coming from. How can I obtain RSA key(or certificate, so I can use Mimikatz) for it?
Edit
I found there is Proprietary Certificate. It seems this is what I need to find, but I still don't know where it is.
Edit: I came across the Proprietary Certificate, but where is private key?
It was located at registry HKLM\SYSTEM\CurrentControlSet\Control\Terminal Server\RCM and is just public key BLOB. Still need to find private key...
Currently I'm looking into registry key Secrets under RCM, but I don't know what are these values right now.
I'm closing this because I found public key BLOB at HKLM\SYSTEM\CurrentControlSet\Control\Terminal Server\RCM\Certificate from registry though I don't know what private key is.
I'm currently working on an old MS-DOS application, which uses DMI to identify the hardware. It worked fine in the past, but it seems to provide invalid data on newer systems (e.g. Skylake). As stated in the spec, we are scanning 0xF0000-0xFFFFF for the "SM" anchor string, this is still working as expected.
But now it seems that the data located at the "Structure table adress" (stored at offset 0x18h in the) are invalid (see dumps below). Tools like dmidecoe deliver correct information (however, it uses GetSystemFirmwareTable() on Windows). What I am doing wrong here?
EDIT (clarify situation)
On an older system I get expected data (dump is done in FreeDOS' debug98 utility) - following come from an IvyBridge system (3rd gen.):
-d F000:04C0
F000:04C0 5F 53 4D 5F 03 1F 02 07-77 00 00 00 00 00 00 00 _SM_....w.......
F000:04D0 5F 44 4D 49 5F E0 6E 04-10 BA 0E 00 17 00 27 00 _DMI_.n.......'.
F000:04E0 1E 66 60 68 00 F0 1F B8-90 D0 83 C0 0F 24 F0 A3 .f`h.........$..
F000:04F0 1D 03 B9 00 E0 2B C8 79-02 33 C9 89 0E 1F 03 33 .....+.y.3.....3
F000:0500 C0 66 2E 8B 1E 63 00 66-83 FB 00 74 0B 66 81 FB .f...c.f...t.f..
F000:0510 00 00 0E 00 72 02 8B C3-A3 19 03 F7 D0 A3 1B 03 ....r...........
F000:0520 66 61 1F C3 00 1E 50 68-00 F0 1F 0B DB 74 28 F7 fa....Ph.....t(.
F000:0530 C3 80 00 74 1C 2E 80 3E-24 05 00 75 43 83 F9 3E ...t...>$..uC..>
-d E000:BA10
E000:BA10 00 18 00 00 01 02 00 F0-03 7F 80 98 89 3F 01 00 .............?..
E000:BA20 00 00 03 0D 04 06 FF FF-41 6D 65 72 69 63 61 6E ........American
E000:BA30 20 4D 65 67 61 74 72 65-6E 64 73 20 49 6E 63 2E Megatrends Inc.
E000:BA40 00 42 51 37 37 52 31 31-31 00 30 37 2F 30 35 2F .BQ77R111.07/05/
E000:BA50 32 30 31 33 00 00 01 1B-01 00 01 02 03 04 00 00 2013............
E000:BA60 01 26 60 24 00 05 00 06-00 07 00 08 00 09 06 05 .&`$............
E000:BA70 06 20 00 20 00 20 00 30-30 30 30 30 31 32 36 36 . . . .000001266
E000:BA80 30 32 34 00 20 00 20 00-00 02 0F 02 00 01 02 03 024. . .........
Newer systems - in this case a Skylake based one (6th gen.) data are different. In the adress the SMI structure points to i do not get the expected data (I expcted to see the BIOS strings, but they are not there):
-d f000:05e0
F000:05E0 5F 53 4D 5F F3 1F 03 00-8C 01 00 00 00 00 00 00 _SM_............
F000:05F0 5F 44 4D 49 5F 15 CE 07-00 90 1D 87 1A 00 30 00 _DMI_.........0.
F000:0600 5F 53 4D 33 5F 4A 18 03-00 00 01 00 CE 07 00 00 _SM3_J..........
F000:0610 00 90 1D 87 00 00 00 00-00 00 00 00 00 00 00 00 ................
F000:0620 1E 66 60 68 00 F0 1F B8-00 C6 83 C0 0F 24 F0 A3 .f`h.........$..
F000:0630 8E 03 B9 00 E0 2B C8 79-02 33 C9 89 0E 90 03 33 .....+.y.3.....3
F000:0640 C0 66 2E 8B 1E 63 00 66-83 FB 00 74 0B 66 81 FB .f...c.f...t.f..
F000:0650 00 00 0E 00 72 02 8B C3-A3 8A 03 F7 D0 A3 8C 03 ....r...........
-d 871d:9000
871D:9000 76 06 D1 E9 73 08 8A 05-A4 88 44 FF 74 08 8B 05 v...s.....D.t...
871D:9010 A5 89 44 FE E2 F8 5F 5E-5D C2 04 00 55 8B EC 4C ..D..._^]...U..L
871D:9020 4C 56 57 83 7E 04 02 73-2D 83 7E 04 02 74 03 E9 LVW.~..s-.~..t..
871D:9030 18 01 8B 46 06 03 06 AC-10 8B F8 50 FF 76 06 FF ...F.......P.v..
871D:9040 16 AE 10 59 59 0B C0 7F-03 E9 FE 00 FF 76 06 57 ...YY........v.W
871D:9050 E8 9D FF E9 F4 00 8B 46-04 48 F7 2E AC 10 8B 56 .......F.H.....V
871D:9060 06 03 D0 8B FA 8B 46 04-D1 E8 F7 2E AC 10 8B 56 ......F........V
871D:9070 06 03 D0 8B F2 57 56 FF-16 AE 10 59 59 0B C0 7E .....WV....YY..~
Your SMBIOS structures are located at physical address 0x871d9000 (as seen from offset f000:0610, or offset x10 from the '_SM3_' anchor string), as Michael Petch points out.
This is a minor point but could be important depending on how your software is constructed. Keep in mind this is a SMBIOS 3.0 conforming structure (per the "_SM3_" anchor string) and that the structure table address can be on any 64-bit address. To ensure your software works in all systems, you should use the _SM3_ structure table address when present and enable your software to read any 64-bit physical address using big-real mode or other mechanism. When the _SM3_ structure is not present, then revert back to your old software flow.
As for why you are just now seeing this, is this the first time you have encountered a data structure that is above 1MB physical address?
I can't finish the authentication phase.
What I am using:
SAM module by NXP
Mifare Desfire PICC
I am following the next steps:
Get PICC SerialNumber (or UID) with GetVersion.
GET VERSION:
Tx: 90 60 00 00 00
Rx: 04 01 01 00 02 18 05 91 AF
GET VERSION 2:
Tx: 90 AF 00 00 00 00
Rx: 04 01 01 00 06 18 05 91 AF
GET VERSION 3:
Tx: 90 AF 00 00 00 00
Rx: 04 65 41 49 65 1B 80 8E 65 58 51 30 46 07 91 00
Get encrypted(RndB) from PICC.
Tx: 90 0A 00 00 01 00 00
Rx: 31 15 1A 19 DB ED CD 5A 91 AF
Send to SAM PICC_SN + ek(RndB).
Tx: 80 41 01 03 0F 80 1B 65 49 41 65 04 31 15 1A 19 DB ED CD 5A
Rx: 61 20
Get from SAM encrypted(RndA + RndB_rotated) + 1st half Session Key
Tx: 00 C0 00 00 20
Rx: F3 10 55 B1 D3 18 91 5B 92 48 16 1F E1 58 D5 CB E9 F3 51 04 41 8A 4E A5 A2 B5 67 CA FF D8 D2 35 90 00
Send PICC encrypted(RndA + RndB_rotated).
Tx: 90 AF 00 00 10 F3 10 55 B1 D3 18 91 5B 92 48 16 1F E1 58 D5 CB 00
Rx: 91 AE
So, this is a guide I have received from my suplier, and i don't have explanations about the apdus used; some i have found them on the internet, some others i guessed.
What I need to know is what does the next command i use:
to SAM module: 80 41 01 03 Lc Data
I need to know what encryption it deploys, why it needs PICC's UID (is this the IV), how can i know RndB, and what is expecting the PICC to end the authentication.
Thanks
Pd: Sorry for the text's format, it seems I'm unable to use correctly the tools for posting, everything gets in the same line it's disgusting...
I solved the problem and finished authentication.
The error was that i was requesting RndB encrypted with keyNo = 0, while corresponding key from SAM's key encryption should be keyNo = 2.
So:
--> 90 0A 00 00 01 02 00
<-- 91 B6 08 CE 9F B5 34 3B 91 AF
Carrying on, i finnish authentication:
--> 90 AF 00 00 10 0F DC FA B6 37 5F 30 34 D7 93 2D A1 3D D6 11 10 00
<-- E9 C2 F2 69 FE 38 78 28 91 00
But now I have the next problem. I've authenticated and I can read PICC's data but i'm afraid it's encrypted. I suppose it is encrypted with session key, so I need some apdu command to be sent to SAM, with data and session key, in order to decrypt data retrieved from PICC.
Am I right? if that is... which would be that SAM APDU?
Trying to understand PNG format.
Consider this PNG Image:
The Image is taken from here
In Hex Editor , it looks like this:
89 50 4E 47 0D 0A 1A 0A 00 00 00 0D 49 48 44 52 00 00 00 80 00 00 00 44 08 02 00 00 00
C6 25 AA 3E 00 00 00 C2 49 44 41 54 78 5E ED D4 81 06 C3 30 14 40 D1 B7 34 DD FF FF 6F
B3 74 56 EA 89 12 6C 28 73 E2 AA 34 49 03 87 D6 FE D8 7B 89 BB 52 8D 3B 87 FE 01 00 80
00 00 10 00 00 02 00 40 00 00 08 00 00 01 00 20 00 00 04 00 80 00 00 10 00 00 02 00 40
00 00 08 00 00 01 00 20 00 00 00 D4 5E 6A 64 4B 94 F5 98 7C D1 F4 92 5C 5C 3E CF 9C 3F
73 71 58 5F AF 8B 79 5B EE 96 B6 47 EB F1 EA D1 CE B6 E3 75 3B E6 B9 95 8D C7 CE 03 39
C9 AF C6 33 93 7B 66 37 CF AB BF F9 C9 2F 08 80 00 00 10 00 00 02 00 40 00 00 08 00 00
01 00 20 00 00 04 00 80 00 00 10 00 00 02 00 40 00 00 08 00 00 01 00 20 00 00 8C 37 DB
68 03 20 FB ED 96 65 00 00 00 00 49 45 4E 44 AE 42 60 82
Equivalent characters:
‰PNG........IHDR...€...D.....Æ%ª>...ÂIDATx^íÔ..Ã0.#Ñ·4Ýÿÿo³tVê‰.l(sâª4I.‡ÖþØ{‰
»R.;‡þ..€.......#....... ....€.......#....... ...Ô^jdK”õ˜|Ñô’\\>Ïœ?sqX_¯
‹y[î–¶GëñêÑζãu;湕.ÇÎ.9ɯÆ3“{f7Ï«¿ùÉ/.€.......#....... ....€.......#....... ..Œ7Ûh.
ûí–e....IEND®B`‚
The same is shown in following Screenshot of the HEX Editor:
I am trying to reverse engineer this image to extract the header part and the RGB pixel values. I read about the PNG and also here , and so far I have noted the following about this Image:
The IHDR chunk must appear FIRST. It contains:
Width: 4 bytes
Height: 4 bytes
Bit depth: 1 byte
Color type: 1 byte
Compression method: 1 byte
Filter method: 1 byte
Interlace method: 1 byte
Below I am starting reading the HEX Data in sequence:
1- First 8-bytes: This is the 8-Byte signature
89 50 4E 47 0D 0A 1A 0A
Equivalently this is : %PNG as can be seen in HEX Editor
A valid PNG image must contain an IHDR chunk, one or more IDAT chunks, and an IEND chunk.
2- Chunk: Length
00 00 00 0D
3-Chunk: Chunk Type
49 48 44 52
Which is IHDR.
http://www.w3.org/TR/PNG-Chunks.html
4- Chunk: Width of the Image (in Decimal 128)
00 00 00 80
5- Chunk: Height of the image (in Decimal 68)
00 00 00 44
6- Chunk: BIT DEPTH (1 byte )
08
7- Chunk: Color Type
02
8- Compression method
00
9- Filter method:
00
10- Interlace method:
00
11- What is the following data?
C6 25 AA 3E 00 00 00 C2
12-- IDAT
49 44 41 54
13- What is this data (after IDAT):
78 5E ED D4 81 06 C3 30 14 40 D1 B7 34 DD FF FF 6F B3 74 56 EA 89 12 6C 28 73 E2 AA 34 49 03 87 D6 FE D8 7B 89 BB 52 8D 3B 87 FE 01 00 80 00 00 10 00 00 02 00 40 00 00 08 00 00 01 00 20 00 00 04 00 80 00 00 10 00 00 02 00 40 00 00 08 00 00 01 00 20 00 00 00 D4 5E 6A 64 4B 94 F5 98 7C D1 F4 92 5C 5C 3E CF 9C 3F 73 71 58 5F AF 8B 79 5B EE 96 B6 47 EB F1 EA D1 CE B6 E3 75 3B E6 B9 95 8D C7 CE 03 39 C9 AF C6 33 93 7B 66 37 CF AB BF F9 C9 2F 08 80 00 00 10 00 00 02 00 40 00 00 08 00 00 01 00 20 00 00 04 00 80 00 00 10 00 00 02 00 40 00 00 08 00 00 01 00 20 00 00 8C 37 DB 68 03 20 FB ED 96 65 00 00 00 00
14- IEND:
49 45 4E 44
15- Last 4 bytes
AE 42 60 82
What are these ?
Can some help me understand, points 11, 13 and 15 above? And where are the Pixel values? The Image is having (128 x 68 pixels)
Purpose of knowing these details:
Once I know these details, I will generate my own 16 bit PNG image. I already have pixel values, so my job would be to introduce headers etc.
I dont know if there is software that can perform this job.
UPDATE
I understand now because of compression, I would not be able to locate the pixel values.
I got the idea that I can write a file in OpenCV and save it as png. Well now my direct question is: I have a binary file having gray-scale 16 bit-pixel values. Can I write this in OpenCV as 16 bit PNG ?
Although it might be interesting to learn about what PNG Images actually are, and how the image is actually represented in the file, you don't need to know this to generate a PNG file.
Note that PNG uses lossless compression, which means you won't get two bytes per pixel.
You can generate your image in a program and output it in PNG format using many of the libraries that there are out there.
For example, you can make your image in OpenCV and then output it with imWrite. One of the parameters can make it output to a PNG.
If you have the gray-scale 16-bit pixel values, then you can put them into a Mat.
Then convert that to an IplImage: Converting cv::Mat to IplImage*
Then you can output it to a file.
Just for completeness (eboix's answer is right on the spot)
11- What is the following data?
C6 25 AA 3E 00 00 00 C2
Each chunk ends with a CRC (4 bytes), and starts with 4 bytes that tell its length.
So, C6 25 AA 3E is the CRC of the previous chunk (IHDR) and 00 00 00 C2 (194) is the length of the following (IDAT) chunk.
In the same way, the last 4 bytes is the CRC of the IEND chunk.
I did not look too carefully but from looking at the structure...
Q11.
C6 25 AA 3E = CRC32
00 00 00 C2 = Size of next chunk
Q13.
check the png spec's you referred to earlier that looks like the IDAT chunk you allready know the compression applied to it!
Q15.
AE 42 60 82 = CRC32