In the registry there is one (or more) key depending how many monitors you have HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Enum\DISPLAY\DEL404C{Some Unique ID}\Device Parameters\EDID which is a REG_BINARY key. In my case this is :
00 ff ff ff ff ff ff 00 4c 2d 6f 03 39 31 59 4d
07 12 01 03 0e 29 1a 78 2a 80 c5 a6 57 49 9b 23
12 50 54 bf ef 80 95 00 95 0f 81 80 81 40 71 4f
01 01 01 01 01 01 9a 29 a0 d0 51 84 22 30 50 98
36 00 ac ff 10 00 00 1c 00 00 00 fd 00 38 4b 1e
51 0e 00 0a 20 20 20 20 20 20 00 00 00 fc 00 53
79 6e 63 4d 61 73 74 65 72 0a 20 20 00 00 00 ff
00 48 56 44 51 32 30 36 37 37 37 0a 20 20 00 ef
My question is how can I read only model of monitor ("SyncMaster" for example) and not all of the information using C or C++?
The format of EDID is described here: http://en.wikipedia.org/wiki/Extended_display_identification_data
What you're interested in here is the descriptor blocks of the EDID, which are found in the byte ranges 54-71, 72-89, 90-107, and 108-125. Here's those four blocks in your EDID:
#1: 9a29 a0d0 5184 2230 5098 3600 acff 1000 00
#2: 0000 00fd 0038 4b1e 510e 000a 2020 2020 20
#3: 0000 00fc 0053 796e 634d 6173 7465 720a 20
#4: 0000 00ff 0048 5644 5132 3036 3737 370a 00
You can identify the descriptor containing the monitor name because the first three bytes are all zero (so it isn't a detailed timing descriptor), and the fourth one byte FC (indicating the type). The fifth byte and beyond contain the name, which is here:
5379 6e63 4d61 7374 6572 0a20 SyncMaster..
So, in short: Check at offsets 54, 72, 90, and 108 for the sequence 00 00 00 FC; if you find a match, the monitor name is the next 12 bytes.
Related
I am using OpenVINO 2020.3.194, with Windows10 x64 and VS2017.I can run the Intel C++ examples, but when I use the inference in my application, I got lots of memory leaks at the exit. Here are some of them:
{2395} normal block at 0x00000260E6A3BCF0, 48 bytes long.
Data: <google.protobuf.> 67 6F 6F 67 6C 65 2E 70 72 6F 74 6F 62 75 66 2E
{2394} normal block at 0x00000260E5958350, 16 bytes long.
Data: < ` > 20 01 A4 E6 60 02 00 00 00 00 00 00 00 00 00 00
{2393} normal block at 0x00000260E6A40100, 88 bytes long.
Data: < ` ` > B0 E4 A2 E6 60 02 00 00 10 E4 A2 E6 60 02 00 00
{2388} normal block at 0x00000260E6A41FB0, 32 bytes long.
Data: <google.protobuf.> 67 6F 6F 67 6C 65 2E 70 72 6F 74 6F 62 75 66 2E
{2387} normal block at 0x00000260E5958170, 16 bytes long.
Data: < ` > 80 00 A4 E6 60 02 00 00 00 00 00 00 00 00 00 00
{2386} normal block at 0x00000260E6A40060, 88 bytes long.
Data: < w ` ` > A0 77 BC E5 60 02 00 00 20 09 A4 E6 60 02 00 00
{2381} normal block at 0x00000260E6A3BC80, 48 bytes long.
Data: <google.protobuf.> 67 6F 6F 67 6C 65 2E 70 72 6F 74 6F 62 75 66 2E
I suspect that happens because I am using Unicode characters set and shared MFC DLL, while the examples compile with MBCS. How I can solve it?
I suggest you try running inference using OpenVINO 2020.4.
Do let us know if this works for you.
I made a program using ffmpeg libraries that converts an audio file to a wav file. Except the only problem is that it doesn't create a 44-byte header. When input the file into Kaldi Speech Recognition, it produces the error:
ERROR (online2-wav-nnet2-latgen-faster:Read4ByteTag():wave-reader.cc:74) WaveData: expected 4-byte chunk-name, got read errror
I ran the file thru shntool and it reports a 78-byte header. Is there anyway I can get the standard 44-byte header using ffmpeg libraries?
FFmpeg inserts some metadata about the encoder into the header file. Here is the hexdump of the header before the fix:
00000000 52 49 46 46 06 90 00 00 57 41 56 45 66 6d 74 20 |RIFF....WAVEfmt |
00000010 10 00 00 00 01 00 01 00 40 1f 00 00 80 3e 00 00 |........#....>..|
00000020 02 00 10 00 4c 49 53 54 1a 00 00 00 49 4e 46 4f |....LIST....INFO|
00000030 49 53 46 54 0e 00 00 00 4c 61 76 66 35 36 2e 33 |ISFT....Lavf56.3|
00000040 36 2e 31 30 30 00 64 61 74 61 c0 8f 00 00 00 00 |6.100.data......|
as you can see Lavf56.36.100 is the encoder in the header. Here is the portion of code that I used to get rid of it.
std::cout<<"------------------BEFORE-----------------------"<<std::endl;
std::cout<< av_dict_count ( (*ofmt_ctx)->metadata) <<std::endl;
std::cout<<"-------------------------------------------"<<std::endl;
if(av_dict_set(&(*ofmt_ctx)->metadata,"ISFT",NULL, AV_DICT_IGNORE_SUFFIX)){
std::cerr<<"Nope it, didn't work :("<<std::endl;
}
ret = avformat_write_header(*ofmt_ctx,&(*ofmt_ctx)->metadata );
if (ret < 0) {
std::cout<<"-------------------------------------------"<<std::endl;
av_log(NULL, AV_LOG_ERROR, "Error occurred when writing header to file\n");
return ret;
}
std::cout<<"------------------AFTER-----------------------"<<std::endl;
std::cout<< av_dict_count ( (*ofmt_ctx)->metadata) <<std::endl;
std::cout<<"-------------------------------------------"<<std::endl;
Here is the hexdump afterwards:
00000000 52 49 46 46 e4 8f 00 00 57 41 56 45 66 6d 74 20 |RIFF....WAVEfmt |
00000010 10 00 00 00 01 00 01 00 40 1f 00 00 80 3e 00 00 |........#....>..|
00000020 02 00 10 00 64 61 74 61 c0 8f 00 00 00 00 00 00 |....data........|
00000030 00 00 00 00 00 00 00 00 ff ff 00 00 00 00 00 00 |................|
shntool now report 44-bytes
(NOTE:ofmt_ctx was a ** in this function that I made, hence why referencing the metadata dictionary as &(*ofmt_ctx)->metadata)
I have to write a small authentication server that, when a client connects, sends a RSA public exponent and modulus to it, and the client returns the username and password encrypted with them. The part where I get the exponent and modulus works fine, the client receives them, and sends the encoded block back; that's not a problem. The problem is in decoding it: the RSA Decryptor always returns 0 as MaxPlaintextLength, and no matter how many different ways I try to move the data around, I can't get it to work.
I'm providing hex dumps of the different parts after the code.
// These are static in the namespace to keep the key loaded.
static CryptoPP::AutoSeededRandomPool rsaRng;
static CryptoPP::InvertibleRSAFunction rsaParameters;
static CryptoPP::RSA::PrivateKey rsaPrivateKey(rsaParameters);
// This is done when the client connects.
{
rsaPrivateKey.GenerateRandomWithKeySize(rsaRng, 1024);
// This is where GetPublicExponent() and GetModulus() are
// encoded and sent to the client. They are sent correctly
// and I receive the encrypted data in a char buffer.
}
// This runs when I receive the encrypted data, which starts
// at &dataBuffer[7] and has a length of rsaPayloadLen.
{
int rsaPlainSize;
byte *rsaBuffer;
rsaBuffer = new byte[rsaPayloadLen];
memcpy(&rsaBuffer[0], &dataBuffer[7], rsaPayloadLen);
CryptoPP::SecByteBlock rsaCypher(rsaBuffer, rsaPayloadLen);
CryptoPP::RSAES_OAEP_SHA_Decryptor rsaDecryptor(rsaPrivateKey);
// At this point I inspected rsaCypher and it does contain the
// expected binary data, and rsaCypher.size() returns 256.
rsaPlainSize = rsaDecryptor.MaxPlaintextLength(rsaCypher.size());
// rsaPlainSize is 0. I have no idea why!
}
Hex dumps of everything at the time of calling MaxPlaintextLength:
rsaPrivateKey.GetPublicExponent():
000000: 11 .
rsaPrivateKey.GetPrivateExponent():
000000: 10 7a fd fd 9e a9 72 8c c3 5d 5b 80 e8 f4 6f bc .z....r..][...o.
000010: bc 6a 7a 51 4f 9f af d3 e4 76 a5 4a 9d fe 17 37 .jzQO....v.J...7
000020: 03 cf 82 24 33 e2 a0 d8 97 26 0a 6b ac 9d b1 de ...$3....&.k....
000030: 39 d5 3a 93 aa 65 66 be 17 43 3c 00 20 77 68 0a 9.:..ef..C<. wh.
000040: ac 2f 77 1e b8 c4 7f 64 52 54 7c 17 54 b6 e6 a4 ./w....dRT|.T...
000050: 95 49 60 7b 7b 16 6a 41 72 54 03 a2 2d 3a 80 8b .I`{{.jArT..-:..
000060: aa 74 fa 77 22 5d 0a d9 81 b2 b2 48 01 db 43 e8 .t.w"].....H..C.
000070: 16 1c c4 c3 a6 bf 45 7e 90 d3 6a 37 10 40 9f 71 ......E~..j7.#.q
rsaPrivateKey.GetModulus():
000000: d2 20 26 61 a6 f0 74 82 ba e6 4e ab 9a 2c 90 a6 . &a..t...N..,..
000010: 62 4d 97 8c b7 34 01 cd a0 e8 bb 77 5e 67 a7 fd bM...4.....w^g..
000020: 70 95 bb 4d 95 89 82 c9 87 25 04 dc d8 da 9b d1 p..M.....%......
000030: 61 5e aa da bc 8c dd f7 a8 99 3d 01 9d f2 6e 89 a^........=...n.
000040: e4 75 ec 91 31 e9 86 f4 da 43 4a ca a4 66 6b 04 .u..1....CJ..fk.
000050: c2 c9 a1 18 1d fa 81 b0 6e ef a5 13 04 44 88 89 ........n....D..
000060: 42 41 be 9c 7c 77 75 96 50 07 70 ad eb 60 e5 05 BA..|wu.P.p..`..
000070: aa a8 d8 27 03 28 cf bb c7 f5 cb 0d b3 b3 96 7f ...'.(..........
rsaPrivateKey.GetPrime1():
000000: d7 9e af ac e4 04 42 e4 58 9c 39 19 0e 56 7c ef ......B.X.9..V|.
000010: b3 bf b6 26 73 25 d8 ab d7 5e d1 e0 56 49 ae 66 ...&s%...^..VI.f
000020: c4 d8 81 bc d0 be c2 ef f4 6a 09 72 ef 72 35 7e .........j.r.r5~
000030: 15 f4 f9 3b f8 be f9 3a a1 0d 3e d0 eb c8 34 11 ...;...:..>...4.
rsaPrivateKey.GetPrime2():
000000: f9 7a 0e 1c 9a 1b eb d1 67 f1 e3 88 1d f3 f1 62 .z......g......b
000010: 9f a2 5c cb 49 76 de 42 25 e1 a4 de ed 50 f3 2d ..\.Iv.B%....P.-
000020: c0 15 c3 70 b5 96 68 51 25 f7 06 24 e4 43 0d b8 ...p..hQ%..$.C..
000030: 7a c5 12 2c 7c 63 20 73 70 61 01 fe b8 b3 71 8f z..,|c spa....q.
Plain text buffer that was encrypted:
000000: 73 74 72 69 6e 67 62 75 66 66 65 72 00 00 00 00 stringbuffer....
000010: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000020: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000030: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000040: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000050: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ..............
rsaCypher:
000000: 0e 9e bd 34 10 16 98 a5 b8 e4 0c 9b 4f 23 71 6d ...4........O#qm
000010: af d6 e8 c1 4d 97 b9 32 cb 25 eb 01 fe 4f 5c 79 ....M..2.%...O\y
000020: 2d d8 32 c4 4f fa e9 2e 58 dd fd 37 7f 08 97 d8 -.2.O...X..7....
000030: 95 bb 6f 04 46 fa 83 77 05 01 43 75 ca be b4 4a ..o.F..w..Cu...J
000040: 60 f9 e7 4a 91 3d bc ac fb e9 41 f3 9d b7 df d3 `..J.=....A.....
000050: a7 03 80 3a 7f 35 98 46 ca 06 b1 f3 d1 7b 56 83 ...:.5.F.....{V.
000060: 1b 00 7d 97 59 39 be 46 d5 cf 6d 2c b3 a7 8e 30 ..}.Y9.F..m,...0
000070: 39 ca ca d5 59 a2 71 43 e7 7e 75 b3 3c d6 a3 a5 9...Y.qC.~u.<...
000080: aa 89 e3 e9 32 e1 a9 c1 a5 a8 f5 66 be 7f c9 ba ....2......f....
000090: 65 35 0f 61 a0 d4 fa c7 ac 8e 28 7c 39 26 3f 01 e5.a......(|9&?.
0000a0: 34 ad 82 69 5e c4 ab 92 48 47 42 04 02 48 79 c4 4..i^...HGB..Hy.
0000b0: 39 6e f2 2c 7c 19 71 34 36 38 1c eb c1 f2 33 f0 9n.,|.q468....3.
0000c0: 49 b9 7e bb c3 16 ed d7 f7 3e 10 a7 cc 2b 8c 31 I.~......>...+.1
0000d0: f1 17 c7 a5 49 ce dd a3 c6 e2 9c 3c 2f 37 e4 97 ....I......</7..
0000e0: ac b7 24 17 b3 f8 75 6f 2a 85 cb 23 7a e1 77 72 ..$...uo*..#z.wr
0000f0: 02 0b 90 28 9b 9b ff 5d 6f 9b 11 11 d3 8b dd 4b ...(...]o......K
rsaCypher.size(): 256
rsaDecryptor.MaxPlaintextLength(rsaCypher.size()): 0
I'm really completely at a loss and the only reason I've spent the whole weekend fighting this is because I'm going to need CryptoPP later for Blowfish and Adler32, otherwise I would have just used another RSA library.
RSAES_OAEP_SHA_Decryptor MaxPlaintextLength returning zero...
Your calculation of rsaPayloadLen is probably not correct. I would suggest you check the size of the ciphertext before you transmit it and after you recover it. I think its size is rsaPayloadLen in you code above.
From the comments in cryptlib.h:
//! maximum length of plaintext for a given ciphertext length
/*! \note This function returns 0 if ciphertextLength is not valid (too long or too short). */
virtual size_t MaxPlaintextLength(size_t ciphertextLength) const =0;
I cannot duplicate when I take plain text, encrypt it, and then pass ciphertext.size() into MaxPlaintextLength. I can duplicate if I pass ciphertext.size() - 1 into MaxPlaintextLength.
As a work around to get you a little further along, you might try using FixedMaxPlaintextLength(). It takes no arguments and only depends upon the public key length. You can use it to find the largest buffer based on the key. But you could fail on decryption.
You should probably show us your encryption code, too. You might not be sizing a buffer correctly.
The whole problem was the key size. The doc said I should use a 1024-bit key, but the other program was expecting a 1152-byte key; apparently that change was never documented. Everything works fine after changing a single line:
rsaPrivateKey.GenerateRandomWithKeySize(rsaRng, 1152);
Leaving this answer here in case someone else ever has this problem. Check the key sizes, and don't trust the documentation!
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Closed 11 years ago.
I've got a program that:
first prints arguments from the command line, with pointers to these arguments, addresses of these pointers and addresses of arguments
then does something similar for the environment variables (however there is a small bug)
after that, all the stack's content is printed starting from buffer "tekst" and going towards the beginning of stack.
At this point, as expected, the program crashes.
The content of the stack is printed as lines:
line number
address in memory
16 consecutive bytes in hexadecimal form
16 consecutive bytes in character form.
Here's a sample output:
ubuntu#ubuntu:~/Desktop$ ./main first_arg second_arg third_arg
argv[0]./main bfd445d1 0xbfd441f4
argv[1]first_arg bfd445d8 0xbfd441f8
argv[2]second_arg bfd445e2 0xbfd441fc
argv[3]third_arg bfd445ed 0xbfd44200
env[0]ORBIT_SOCKETDIR=/tmp/orbit-ubuntu bfd445f7 0xbfd44208
env[1]SSH_AGENT_PID=4294 bfd445f7 0xbfd4420c
env[2]TERM=xterm bfd445f7 0xbfd44210
env[3]SHELL=/bin/bash bfd445f7 0xbfd44214
env[4]XDG_SESSION_COOKIE=b16da184017a223e33dfbc7700000025-1316532826.4763-1998941864 bfd445f7 0xbfd44218
env[5]WINDOWID=60817444 bfd445f7 0xbfd4421c
env[6]GNOME_KEYRING_CONTROL=/tmp/keyring-PF9cWq bfd445f7 0xbfd44220
env[7]GTK_MODULES=canberra-gtk-module bfd445f7 0xbfd44224
env[8]USER=ubuntu bfd445f7 0xbfd44228
env[9]LS_COLORS=rs=0:di=01;34:ln=01;36:mh=00:pi=40;33:so=01;35:do=01;35:bd=40;33;01:cd=40;33;01:or=40;31;01:su=37;41:sg=30;43:ca=30;41:tw=30;42:ow=34;42:st=37;44:ex=01;32:*.tar=01;31:*.tgz=01;31:*.arj=01;31:*.taz=01;31:*.lzh=01;31:*.lzma=01;31:*.tlz=01;31:*.txz=01;31:*.zip=01;31:*.z=01;31:*.Z=01;31:*.dz=01;31:*.gz=01;31:*.lz=01;31:*.xz=01;31:*.bz2=01;31:*.bz=01;31:*.tbz=01;31:*.tbz2=01;31:*.tz=01;31:*.deb=01;31:*.rpm=01;31:*.jar=01;31:*.rar=01;31:*.ace=01;31:*.zoo=01;31:*.cpio=01;31:*.7z=01;31:*.rz=01;31:*.jpg=01;35:*.jpeg=01;35:*.gif=01;35:*.bmp=01;35:*.pbm=01;35:*.pgm=01;35:*.ppm=01;35:*.tga=01;35:*.xbm=01;35:*.xpm=01;35:*.tif=01;35:*.tiff=01;35:*.png=01;35:*.svg=01;35:*.svgz=01;35:*.mng=01;35:*.pcx=01;35:*.mov=01;35:*.mpg=01;35:*.mpeg=01;35:*.m2v=01;35:*.mkv=01;35:*.ogm=01;35:*.mp4=01;35:*.m4v=01;35:*.mp4v=01;35:*.vob=01;35:*.qt=01;35:*.nuv=01;35:*.wmv=01;35:*.asf=01;35:*.rm=01;35:*.rmvb=01;35:*.flc=01;35:*.avi=01;35:*.fli=01;35:*.flv=01;35:*.gl=01;35:*.dl=01;35:*.xcf=01;35:*.xwd=01;35:*.yuv=01;35:*.cgm=01;35:*.emf=01;35:*.axv=01;35:*.anx=01;35:*.ogv=01;35:*.ogx=01;35:*.aac=00;36:*.au=00;36:*.flac=00;36:*.mid=00;36:*.midi=00;36:*.mka=00;36:*.mp3=00;36:*.mpc=00;36:*.ogg=00;36:*.ra=00;36:*.wav=00;36:*.axa=00;36:*.oga=00;36:*.spx=00;36:*.xspf=00;36: bfd445f7 0xbfd4422c
env[a]SSH_AUTH_SOCK=/tmp/keyring-PF9cWq/ssh bfd445f7 0xbfd44230
env[b]SESSION_MANAGER=local/ubuntu:#/tmp/.ICE-unix/4253,unix/ubuntu:/tmp/.ICE-unix/4253 bfd445f7 0xbfd44234
env[c]USERNAME=ubuntu bfd445f7 0xbfd44238
env[d]DEFAULTS_PATH=/usr/share/gconf/gnome.default.path bfd445f7 0xbfd4423c
env[e]XDG_CONFIG_DIRS=/etc/xdg/xdg-gnome:/etc/xdg bfd445f7 0xbfd44240
env[f]PATH=/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games bfd445f7 0xbfd44244
env[10]DESKTOP_SESSION=gnome bfd445f7 0xbfd44248
env[11]PWD=/home/ubuntu/Desktop bfd445f7 0xbfd4424c
env[12]GDM_KEYBOARD_LAYOUT=us bfd445f7 0xbfd44250
env[13]LANG=en_US.UTF-8 bfd445f7 0xbfd44254
env[14]GDM_LANG=en_US.UTF-8 bfd445f7 0xbfd44258
env[15]MANDATORY_PATH=/usr/share/gconf/gnome.mandatory.path bfd445f7 0xbfd4425c
env[16]UBUNTU_MENUPROXY=libappmenu.so bfd445f7 0xbfd44260
env[17]COMPIZ_CONFIG_PROFILE=ubuntu bfd445f7 0xbfd44264
env[18]GDMSESSION=gnome bfd445f7 0xbfd44268
env[19]SHLVL=1 bfd445f7 0xbfd4426c
env[1a]HOME=/home/ubuntu bfd445f7 0xbfd44270
env[1b]LANGUAGE=en_US:en bfd445f7 0xbfd44274
env[1c]GNOME_DESKTOP_SESSION_ID=this-is-deprecated bfd445f7 0xbfd44278
env[1d]LOGNAME=ubuntu bfd445f7 0xbfd4427c
env[1e]XDG_DATA_DIRS=/usr/share/gnome:/usr/local/share/:/usr/share/ bfd445f7 0xbfd44280
env[1f]DBUS_SESSION_BUS_ADDRESS=unix:abstract=/tmp/dbus-DTbn1HNZFU,guid=3b2a0c2f55352a2d70678ffc000000a3 bfd445f7 0xbfd44284
env[20]LESSOPEN=| /usr/bin/lesspipe %s bfd445f7 0xbfd44288
env[21]WINDOWPATH=7 bfd445f7 0xbfd4428c
env[22]DISPLAY=:0.0 bfd445f7 0xbfd44290
env[23]LESSCLOSE=/usr/bin/lesspipe %s %s bfd445f7 0xbfd44294
env[24]RUNNING_UNDER_GDM=yes bfd445f7 0xbfd44298
env[25]COLORTERM=gnome-terminal bfd445f7 0xbfd4429c
env[26]XAUTHORITY=/var/run/gdm/auth-for-ubuntu-Za0Xwx/database bfd445f7 0xbfd442a0
env[27]_=./main bfd445f7 0xbfd442a4
env[28]OLDPWD=/home/ubuntu bfd445f7 0xbfd442a8
001 0xbfd4411e 48 45 52 45 5f 49 53 5f 53 54 41 52 54 00 00 f0 HERE_IS_START???
002 0xbfd4412e fc e2 35 ad 28 00 f4 7f 3b 00 00 00 00 00 00 00 ??5?(???;???????
003 0xbfd4413e 00 00 60 8d 04 08 00 00 00 00 c8 41 d4 bf 37 1e ??`????????A??7?
004 0xbfd4414e 27 00 04 00 00 00 f4 41 d4 bf 08 42 d4 bf 14 44 '??????A???B???D
005 0xbfd4415e 85 00 ff ff ff ff f4 9f b7 00 f4 84 04 08 01 00 ????????????????
006 0xbfd4416e 00 00 b0 41 d4 bf 31 aa b6 00 d0 aa b7 00 30 86 ???A??1???????0?
007 0xbfd4417e 8c b7 01 00 00 00 f4 7f 3b 00 00 00 00 00 00 00 ????????;???????
008 0xbfd4418e 00 00 c8 41 d4 bf c1 93 97 8b be ca 2e 6d 00 00 ???A????????.m??
009 0xbfd4419e 00 00 00 00 00 00 00 00 00 00 04 00 00 00 60 87 ??????????????`?
010 0xbfd441ae 04 08 00 00 00 00 40 0c b7 00 5b 1d 27 00 f4 9f ??????#???[?'???
011 0xbfd441be b7 00 04 00 00 00 60 87 04 08 00 00 00 00 81 87 ??????`?????????
012 0xbfd441ce 04 08 14 88 04 08 04 00 00 00 f4 41 d4 bf 60 8d ???????????A??`?
013 0xbfd441de 04 08 c0 8d 04 08 50 ba b6 00 ec 41 d4 bf 18 a9 ??????P????A????
014 0xbfd441ee b7 00 04 00 00 00 d1 45 d4 bf d8 45 d4 bf e2 45 ???????E???E???E
015 0xbfd441fe d4 bf ed 45 d4 bf 00 00 00 00 f7 45 d4 bf 19 46 ???E???????E???F
016 0xbfd4420e d4 bf 2c 46 d4 bf 37 46 d4 bf 47 46 d4 bf 96 46 ??,F??7F??GF???F
017 0xbfd4421e d4 bf a8 46 d4 bf d2 46 d4 bf f2 46 d4 bf fe 46 ???F???F???F???F
018 0xbfd4422e d4 bf ee 4b d4 bf 14 4c d4 bf 66 4c d4 bf 76 4c ???K???L??fL??vL
019 0xbfd4423e d4 bf a8 4c d4 bf d4 4c d4 bf 21 4d d4 bf 37 4d ???L???L??!M??7M
020 0xbfd4424e d4 bf 50 4d d4 bf 67 4d d4 bf 78 4d d4 bf 8d 4d ??PM??gM??xM???M
021 0xbfd4425e d4 bf c2 4d d4 bf e1 4d d4 bf fe 4d d4 bf 0f 4e ???M???M???M???N
022 0xbfd4426e d4 bf 17 4e d4 bf 29 4e d4 bf 3b 4e d4 bf 67 4e ???N??)N??;N??gN
023 0xbfd4427e d4 bf 76 4e d4 bf b3 4e d4 bf 15 4f d4 bf 35 4f ??vN???N???O??5O
024 0xbfd4428e d4 bf 42 4f d4 bf 4f 4f d4 bf 71 4f d4 bf 87 4f ??BO??OO??qO???O
025 0xbfd4429e d4 bf a0 4f d4 bf d8 4f d4 bf e1 4f d4 bf 00 00 ???O???O???O????
026 0xbfd442ae 00 00 20 00 00 00 14 44 85 00 21 00 00 00 00 40 ?? ????D??!????#
027 0xbfd442be 85 00 10 00 00 00 bf f3 8b 07 06 00 00 00 00 10 ????????????????
028 0xbfd442ce 00 00 11 00 00 00 64 00 00 00 03 00 00 00 34 80 ??????d???????4?
029 0xbfd442de 04 08 04 00 00 00 20 00 00 00 05 00 00 00 09 00 ?????? ?????????
030 0xbfd442ee 00 00 07 00 00 00 00 d0 b5 00 08 00 00 00 00 00 ????????????????
031 0xbfd442fe 00 00 09 00 00 00 60 87 04 08 0b 00 00 00 e7 03 ??????`?????????
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235 0xbfd44fbe 6f 72 2d 75 62 75 6e 74 75 2d 5a 61 30 58 77 78 or-ubuntu-Za0Xwx
236 0xbfd44fce 2f 64 61 74 61 62 61 73 65 00 5f 3d 2e 2f 6d 61 /database?_=./ma
237 0xbfd44fde 69 6e 00 4f 4c 44 50 57 44 3d 2f 68 6f 6d 65 2f in?OLDPWD=/home/
238 0xbfd44fee 75 62 75 6e 74 75 00 2e 2f 6d 61 69 6e 00 00 00 ubuntu?./main???
239 0xbfd44ffe 00 00 Segmentation fault (core dumped)
At the moment, I have figured out following things (starting from line 001):
001 buffer tekst
014 argc
014-015 pointers to argv[i]
015 4x null byte
015-025 pointers to env[i]
025-026 4x null byte
038-076 many 0 bytes
076-078 strings of command line arguments
078-238 strings of environments variables
238 name of program
238-239 4x null byte
I have two questions:
What is the other content of stack?
What are used these multiple zeroes between lines 038 and 076 for?
For reference, here's the code I'm using:
#include <iostream>
#include <locale>
#include <unistd.h>
int main(int argc, char* argv[], char* env[]) {
char tekst[] = "HERE_IS_START";
for (int i = 0; i < argc; i++) {
std::cout << "argv[" << i << "]" << argv[i] << " " << std::hex <<
(int) argv[i] << " " << &argv[i] << std::endl;
}
int z = 0;
for (char** env_ = env; *env_ != 0; env_++) {
std::cout << "env[" << z++ << "]" << *env_ << " " << std::hex <<
(int) *env << " " << &(*env_) << std::endl;
}
int row = 16;
char* wsk = tekst;
for (int i = 0;; i++) {
if (i + 1 < 10)std::cout << 0;
if (i + 1 < 100)std::cout << 0;
std::cout << std::dec << (i + 1) << " 0x" << std::hex << (int) wsk << " ";
for (int j = 0; j < row; j++) {
if ((unsigned int) (unsigned char) (wsk[j]) < 16)
std::cout << 0;
std::cout << (unsigned int) (unsigned char) (wsk[j]) <<
" " << std::flush;
}
for (int j = 0; j < row; j++) {
if (isprint((char) (wsk[j])))
std::cout << (char) (wsk[j]);
else
std::cout << '?';
}
std::cout << std::endl;
wsk = wsk + row;
}
return 0;
}
This program was compiled and tested in Ubuntu 11.04.
How do you ever expect to break out of that for loop? You don't give it a repeat condition, and you don't have an explicit break or return anywhere inside the loop. Presumably this segmentation fault is caused by going past the end of the stack into memory you're not allowed to access.
There are two things that will put stuff on the stack before main starts: the operating system itself, and the run-time library and startup code generated by your compiler. You aren't expected to use any of this information directly so it is generally not documented.
I'm working on understanding and drawing my own DLL for PDF417 (2d barcodes). Anyhow, the actual drawing of the file is perfect, and in correct boundaries of 32 bits (as monochrome result). At the time of writing the data, the following is a memory dump as copied from C++ Visual Studio memory dump of the pointer to the bmp buffer. Each row is properly allocated to 36 wide before the next row.
Sorry about the wordwrap in the post, but my output was intended to be the same 36 bytes wide as the memory dump so you could better see the distortion.
The current drawing is 273 pixels wide by 12 pixels high, monochrome...
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00 00
00 ab 85 7e d0 29 e8 14 f4 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0a 8e 02 eb 00 00
00 ab 85 7e d0 29 e8 14 f4 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0a 8e 02 eb 00 00
00 ab 85 7e d0 29 e8 14 f4 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0a 8e 02 eb 00 00
00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c db ce 6d 5f be 02 eb 00 00
00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c db ce 6d 5f be 02 eb 00 00
00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c db ce 6d 5f be 02 eb 00 00
Here is the code to WRITE the file out -- verbatim immediately at the time of the memory dump from above
FILE *stream;
if( fopen_s( &stream, cSaveToFile, "w+" ) == 0 )
{
fwrite( &bmfh, 1, (UINT)sizeof(BITMAPFILEHEADER), stream );
fwrite( &bmi, 1, (UINT)sizeof(BITMAPINFO), stream );
fwrite( &RGBWhite, 1, (UINT)sizeof(RGBQUAD), stream );
fwrite( ppvBits, 1, (UINT)bmi.bmiHeader.biSizeImage, stream );
fclose( stream );
}
Here's what ACTUALLY Gets written to the file.
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0d 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00
00 00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0d 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb
00 00 00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0d 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02
eb 00 00 00 ab 85 7e d0 29 e8 14 f4 0d 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0d 0a
8e 02 eb 00 00 00 ab 85 7e d0 29 e8 14 f4 0d 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c
0d 0a 8e 02 eb 00 00 00 ab 85 7e d0 29 e8 14 f4 0d 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79
bc 5c 0d 0a 8e 02 eb 00 00 00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c
db ce 6d 5f be 02 eb 00 00 00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c
db ce 6d 5f be 02 eb 00 00 00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c
db ce 6d 5f be 02 eb 00 00
Notice the start of the distortion with the "0d" in the result from reading the file back in the 4th line, about the 15th byte over... Then, there are a few more staggered around which in total, skew the image off by 9 bytes worth...
Obviously, the drawing portion is working ok as everything remains properly aligned in memory for the 12 lines.
Shouldn't you open the file in a compound mode i.e. writable & binary as in wb+?
Notice the start of the distortion with the "0d"
That's ASCII code for Carriage Return (CR) -- added on some OSes with newline (where a newline is actually a sequence of CR/LF). This should go away once you start writing the output in binary mode.
Your code looks neat otherwise. Cheers!
Your 0x0A (\n) gets converted to DOS format 0x0D0A (\r\n), becouse you're write the file in text mode. Switch to binary mode.
I actually just did a similar thing in java (printing bmp data to a thermal receipt printer). There are a couple of things i want to share with you:
bmp image data != an image format from microsoft. the MS bitmap has about 54 bytes of header information before any image data. (i spent a day or two working on this before I realized the difference)
bmp image data reads left to right, top to bottom, with the most significant bit on the left.
make sure the barcode image has a bitdepth of 1. this means 1 bit = 1 pixel. hexidecimal "ab" is 10101011 in binary, those 8 pixels will be filled in accordingly.
if you have a barcode 36 bytes wide, the barcode resolution is 288 x 12, not 273 x 12. (36 * 8 = 288).
the image data should be 432 bytes in size (12 rows of 36 bytes).
i dont know what this means:
Anyhow, the actual drawing of the file is perfect, and in correct boundaries of 32 bits (as monochrome result).
monochrome means its either 1 color or another. the pixel (think bit) is either filled in or it isnt.
Hope this helps