- file
- xxd
- nm
- readelf
-
Setup Let's m ake a new working directory:
$ mkdir ~/PBA $ cd ~/PBA
And now let's download the example payload file:
$ wget https://practicalbinaryanalysis.com/file/pba-code.tar.gz
Unzip and expand the archive andchange into chapter5
$ tar xzf pba-code.tar.gz $ cd code/chapter5 -
Determine the file type
We'll start by using the
filecommand to determine the filetype:$ file payload payload: ASCII text
Looks like an ASCII file. Let's see what the ASCII content looks like:
$ cat payload H4sIABzY61gAA+xaD3RTVZq/Sf+lFJIof1r+2aenKKh0klJKi4MmJaUvWrTSFlgR0jRN20iadpKX UljXgROKjbUOKuOfWWfFnTlzZs/ZXTln9nTRcTHYERhnZ5c/R2RGV1lFTAFH/DNYoZD9vvvubd57 ... AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAODbbqzb /Q4AWAwA
We can see both lowercase (26) and uppercase (26) characters as well as the
+and/characters for a total character set of 64. Best guess is that it's a base64 encoded file. Let's move the payload to a more descriptive name:$ mv payload payload.base64
And now lets go ahead and decode the base64 encoded file:
$ base64 -d payload.base64 > payload.decodedAnd use
fileagain to determine the filetype of the decoded content:$ file payload.decoded payload.decoded: gzip compressed data, last modified: Mon Apr 10 19:08:12 2017, from Unix, original size 808960
Looks like a gzip file. Let's use the
filecommand to look inside the gzip file:**$ file -z payload.decoded payload.decoded: POSIX tar archive (GNU) (gzip compressed data, last modified: Mon Apr 10 19:08:12 2017, from Unix)
Looks like a tar archive inside the gzip file, making this a .tgz file. Let's move it to something more expressive:
$ mv payload.decoded payload.tgz
-
Examine the gzip'd tar archive
Start by decompressing and extracting the contents of the tar archive:
$ tar xvzf payload.tgz ctf 67b8601
There appears to be 2 files in the archvie. Let's determine the filetypes of these 2 files:
$ file ctf ctf: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=29aeb60bcee44b50d1db3a56911bd1de93cd2030, stripped $ file 67b8601 67b8601: PC bitmap, Windows 3.x format, 512 x 512 x 24Looks like we have an ELF binary named
ctfand a Windows bitmap graphic named67b8601. -
Examine the
ctfELF binary fileLet's start by running the binary. Typically you would not just run this mysterious binary anywhere and would instead do it in a more controlled and ideally air-gapped environment. Because this binary comes from "No Starch Press" as part of their "Practical Binary Analysis* book we will go ahead and run it:
$ ./ctf ./ctf: error while loading shared libraries: lib5ae9b7f.so: cannot open shared object file: No such file or directoryInteresting. Looks like the binary is missing a library when looking for dependencies. Let's take a closer look with the
lddtool and see what libraries it is looking for and which ones it can and can't find:$ ldd ctf linux-vdso.so.1 (0x00007ffeddb8a000) lib5ae9b7f.so => not found libstdc++.so.6 => /usr/lib/x86_64-linux-gnu/libstdc++.so.6 (0x00007fe4e276f000) libgcc_s.so.1 => /lib/x86_64-linux-gnu/libgcc_s.so.1 (0x00007fe4e2755000) libc.so.6 => /lib/x86_64-linux-gnu/libc.so.6 (0x00007fe4e2594000) libm.so.6 => /lib/x86_64-linux-gnu/libm.so.6 (0x00007fe4e2411000) /lib64/ld-linux-x86-64.so.2 (0x00007fe4e2921000)
Notice that we are missing a library named
lib5ae9b7f.so, so we'll likely need to address that. This is not a standard library file, so it likely came with files in the tar archive. GIven this is an ELF binary, let's search for the stringELFin the files we have:$ grep ELF * Binary file 67b8601 matches Binary file ctf matchesLooks like both the ELF binary (which we expected) and the Windows bitmap file contain the string. Let's take a closer look at the bitmap file.
-
Examine the
67b8601Windows bitmap fileLet's start by taking a look at the first 10 lines of the bitmap file:
$ xdd 67b8601 | head 00000000: 424d 3800 0c00 0000 0000 3600 0000 2800 BM8.......6...(. 00000010: 0000 0002 0000 0002 0000 0100 1800 0000 ................ 00000020: 0000 0200 0c00 c01e 0000 c01e 0000 0000 ................ 00000030: 0000 0000 7f45 4c46 0201 0100 0000 0000 .....ELF........ 00000040: 0000 0000 0300 3e00 0100 0000 7009 0000 ......>.....p... 00000050: 0000 0000 4000 0000 0000 0000 7821 0000 [email protected]!.. 00000060: 0000 0000 0000 0000 4000 3800 0700 4000 [email protected]...@. 00000070: 1b00 1a00 0100 0000 0500 0000 0000 0000 ................ 00000080: 0000 0000 0000 0000 0000 0000 0000 0000 ................ 00000090: 0000 0000 f40e 0000 0000 0000 f40e 0000 ................
We can see the
ELFstring is on the 4th line. We want to grab everything from theELFstring forward, and drop everything leading up to it. This means we want to skip a specific number of bytes so we can create a new file that begins with the ELF header.In the output of
xxdabove, every 2 digits represents a single byte, and each line represents 16 bytes. That means the first 3 lines represent 48 bytes, and the 8 characters leading up to7f45(ELF) represent 4 bytes, for a total offset of 52 bytes before we get to theELFstring.Let's pull the ELF header out by skipping the first 52 bytes and then dumping the next 64 bytes into a new file. We have no idea how much to grab, so we'll grab 64 bytes for now and adjust later.
$ dd if=67b8601 \ of=67b8601.elf_header \ skip=52 \ count=64 \ bs=1 64+0 records in 64+0 records out 64 bytes copied, 0.000683955 s, 93.6 kB/sLet's check out the contents of the ELF header file we jsut created:
$ xxd 67b8601.elf_header 00000000: 7f45 4c46 0201 0100 0000 0000 0000 0000 .ELF............ 00000010: 0300 3e00 0100 0000 7009 0000 0000 0000 ..>.....p....... 00000020: 4000 0000 0000 0000 7821 0000 0000 0000 @.......x!...... 00000030: 0000 0000 4000 3800 0700 4000 1b00 1a00 [email protected]...@.....
Looks like we got the right stuff. Our file now begins with the ELF header :)
-
Examine the ELF header file
$ readelf -h 67b8601.elf_header ELF Header: Magic: 7f 45 4c 46 02 01 01 00 00 00 00 00 00 00 00 00 Class: ELF64 Data: 2's complement, little endian Version: 1 (current) OS/ABI: UNIX - System V ABI Version: 0 Type: DYN (Shared object file) Machine: Advanced Micro Devices X86-64 Version: 0x1 Entry point address: 0x970 Start of program headers: 64 (bytes into file) Start of section headers: 8568 (bytes into file) Flags: 0x0 Size of this header: 64 (bytes) Size of program headers: 56 (bytes) Number of program headers: 7 Size of section headers: 64 (bytes) Number of section headers: 27 Section header string table index: 26 readelf: Error: Reading 1728 bytes extends past end of file for section headers readelf: Error: Too many program headers - 0x7 - the file is not that big
Looks like we can read the file, but there are some errors at the bottom because we guessed 64 bytes and were wrong. Let's determine the exact size of the ELF file embedded in the bitmap file.
Start by multiplying the values of the
Size of section headersandNumber of section headersfields and then add the result to theStart of section headersfield. In this case, we can represent our specific formula as64 * 27 + 8568. Let's use theexprcommand to determine the result:$ expr 64 \* 27 + 8568 10296Let's use the result of
10296along with our ELF header offset of 52 bytes to grab the exact bytes we need from the originalctfbinary. Let's also change the output filename to the name of the missing library (lib5ae9b7f.so), given we know we are missing an ELF library:$ dd if=67b8601 \ of=lib5ae9b7f.so \ skip=52 \ count=10296 \ bs=1 10296+0 records in 10296+0 records out 10296 bytes (10 kB, 10 KiB) copied, 0.0466762 s, 221 kB/sLet's take another look with the
readelfcommand and see if we still get those errors:$ readelf -hs lib5ae9b7f.so ELF Header: Magic: 7f 45 4c 46 02 01 01 00 00 00 00 00 00 00 00 00 Class: ELF64 Data: 2's complement, little endian Version: 1 (current) OS/ABI: UNIX - System V ABI Version: 0 Type: DYN (Shared object file) Machine: Advanced Micro Devices X86-64 Version: 0x1 Entry point address: 0x970 Start of program headers: 64 (bytes into file) Start of section headers: 8568 (bytes into file) Flags: 0x0 Size of this header: 64 (bytes) Size of program headers: 56 (bytes) Number of program headers: 7 Size of section headers: 64 (bytes) Number of section headers: 27 Section header string table index: 26 Symbol table '.dynsym' contains 22 entries: Num: Value Size Type Bind Vis Ndx Name 0: 0000000000000000 0 NOTYPE LOCAL DEFAULT UND 1: 00000000000008c0 0 SECTION LOCAL DEFAULT 9 2: 0000000000000000 0 NOTYPE WEAK DEFAULT UND __gmon_start__ 3: 0000000000000000 0 NOTYPE WEAK DEFAULT UND _Jv_RegisterClasses 4: 0000000000000000 0 FUNC GLOBAL DEFAULT UND _ZNSt7__cxx1112basic_stri@GLIBCXX_3.4.21 (2) 5: 0000000000000000 0 FUNC GLOBAL DEFAULT UND malloc@GLIBC_2.2.5 (3) 6: 0000000000000000 0 NOTYPE WEAK DEFAULT UND _ITM_deregisterTMCloneTab 7: 0000000000000000 0 NOTYPE WEAK DEFAULT UND _ITM_registerTMCloneTable 8: 0000000000000000 0 FUNC WEAK DEFAULT UND __cxa_finalize@GLIBC_2.2.5 (3) 9: 0000000000000000 0 FUNC GLOBAL DEFAULT UND __stack_chk_fail@GLIBC_2.4 (4) 10: 0000000000000000 0 FUNC GLOBAL DEFAULT UND _ZSt19__throw_logic_error@GLIBCXX_3.4 (5) 11: 0000000000000000 0 FUNC GLOBAL DEFAULT UND memcpy@GLIBC_2.14 (6) 12: 0000000000000bc0 149 FUNC GLOBAL DEFAULT 12 _Z11rc4_encryptP11rc4_sta 13: 0000000000000cb0 112 FUNC GLOBAL DEFAULT 12 _Z8rc4_initP11rc4_state_t 14: 0000000000202060 0 NOTYPE GLOBAL DEFAULT 24 _end 15: 0000000000202058 0 NOTYPE GLOBAL DEFAULT 23 _edata 16: 0000000000000b40 119 FUNC GLOBAL DEFAULT 12 _Z11rc4_encryptP11rc4_sta 17: 0000000000000c60 5 FUNC GLOBAL DEFAULT 12 _Z11rc4_decryptP11rc4_sta 18: 0000000000202058 0 NOTYPE GLOBAL DEFAULT 24 __bss_start 19: 00000000000008c0 0 FUNC GLOBAL DEFAULT 9 _init 20: 0000000000000c70 59 FUNC GLOBAL DEFAULT 12 _Z11rc4_decryptP11rc4_sta 21: 0000000000000d20 0 FUNC GLOBAL DEFAULT 13 _fini
Excellent! It looks like we cut out the right number of bytes from the right offset and now have a complete ELF library anmed after the missing library we found in the
lddcommand earlier. Let's try thelddcommand again to see if it finds the library:$ LD_LIBRARY_PATH=$(pwd) ldd ctf LD_LIBRARY_PATH=$(pwd) ldd ctf linux-vdso.so.1 (0x00007fffb7d10000) lib5ae9b7f.so => /usr/home/example/PBA/C5/lib5ae9b7f.so (0x00007fb82e09a000) libstdc++.so.6 => /usr/lib/x86_64-linux-gnu/libstdc++.so.6 (0x00007fb82deea000) libgcc_s.so.1 => /lib/x86_64-linux-gnu/libgcc_s.so.1 (0x00007fb82ded0000) libc.so.6 => /lib/x86_64-linux-gnu/libc.so.6 (0x00007fb82dd0f000) libm.so.6 => /lib/x86_64-linux-gnu/libm.so.6 (0x00007fb82db8c000) /lib64/ld-linux-x86-64.so.2 (0x00007fb82e29f000)
Looks like the binary can find all of its libraries now :) Let's try to run it again with this new library file:
$ LD_LIBRARY_PATH=$(pwd) ./ctf $ echo $? 1
Success! The binary can now run but returns no output. Checking the exit code (
$?) reveals an exit code of1. Non-zero exit codes repsent failure, so while the binary can now run it is not completing successfully. Time for further analysis. -
Demangling the function names
Let's take a closer look at the function names of this binary and see if we can better understand what this program is used for. We'll use the
nmcommand to take a look at function names:$ nm lib5ae9b7f.so nm: lib5ae9b7f.so: no symbols
Looks like this binary has been stripped of symbols, so we'll need to pass the
-Dflag tonmin order to show dynamic symbols:$ nm -D lib5ae9b7f.so 0000000000202058 B __bss_start w __cxa_finalize 0000000000202058 D _edata 0000000000202060 B _end 0000000000000d20 T _fini w __gmon_start__ 00000000000008c0 T _init w _ITM_deregisterTMCloneTable w _ITM_registerTMCloneTable w _Jv_RegisterClasses U malloc U memcpy U __stack_chk_fail 0000000000000c60 T _Z11rc4_decryptP11rc4_state_tPhi 0000000000000c70 T _Z11rc4_decryptP11rc4_state_tRNSt7__cxx1112basic_stringIcSt11char_traitsIcESaIcEEE 0000000000000b40 T _Z11rc4_encryptP11rc4_state_tPhi 0000000000000bc0 T _Z11rc4_encryptP11rc4_state_tRNSt7__cxx1112basic_stringIcSt11char_traitsIcESaIcEEE 0000000000000cb0 T _Z8rc4_initP11rc4_state_tPhi U _ZNSt7__cxx1112basic_stringIcSt11char_traitsIcESaIcEE9_M_createERmm U _ZSt19__throw_logic_errorPKcThat helps a bit, but we still have mangled funciton names. Let's pass the
--demangleflag tonmand see fi that cleans things up:$ nm -D --demangle lib5ae9b7f.so 0000000000202058 B __bss_start w __cxa_finalize 0000000000202058 D _edata 0000000000202060 B _end 0000000000000d20 T _fini w __gmon_start__ 00000000000008c0 T _init w _ITM_deregisterTMCloneTable w _ITM_registerTMCloneTable w _Jv_RegisterClasses U malloc U memcpy U __stack_chk_fail 0000000000000c60 T rc4_decrypt(rc4_state_t*, unsigned char*, int) 0000000000000c70 T rc4_decrypt(rc4_state_t*, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&) 0000000000000b40 T rc4_encrypt(rc4_state_t*, unsigned char*, int) 0000000000000bc0 T rc4_encrypt(rc4_state_t*, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&) 0000000000000cb0 T rc4_init(rc4_state_t*, unsigned char*, int) U std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >::_M_create(unsigned long&, unsigned long) U std::__throw_logic_error(char const*)Much better. Its now easier to see that this appears to be an encryption/decryption tool of some sort, likely using the rc4 encryption scheme.
Here's another way to demangle C++ function names from a binary using the
c++filtcommand to turn a mangled function name into a demangled function name:$ c++filt _Z11rc4_decryptP11rc4_state_tPhi rc4_decrypt(rc4_state_t*, unsigned char*, int)
This works well, but requires you to do it for each function name, rather than the
nmcommand that lets you demangle all of them at once. -
Look for intersting strings in the binary
The
stringscommand makes it easy to dump all of the strings where there are N number of contiguous ASCII characters. The default number is 4 but can be tuned using CLI options.$ strings ctf /lib64/ld-linux-x86-64.so.2 ... DEBUG: argv[1] = %s checking '%s' show_me_the_flag ... flag = %s guess again! It's kinda like Louisiana. Or Dagobah. Dagobah - Where Yoda lives! ... GCC: (Ubuntu 5.4.0-6ubuntu1~16.04.4) 5.4.0 20160609 .shstrtab ...
I've omitted all but the most intersting strings to make it easier to see what might be useful.
The
DEBUG: argv[1] = %sstring is especially interesting as it implies that this binary accepts an argument. Then we seechecking '%s'which is probably output to the screen to let the user know it is now checking the argument. The next string isshow_me_the_flagwhich might be the argument value expected by the binary. Let's give it a try:$ LD_LIBRARY_PATH=$(pwd) ./ctf show_me_the_flag checking 'show_me_the_flag' ok $ echo $? 1
This looks like progress :) Right away we see the
checking 'show_me_the_flag'which matches thechecking '%s'string we saw before in the output of thestringscommand. And then we check the exit code and see that it is still non-zero, implying we have more work to do. Let's also try sending the wrong argument and see how it differs, if at all:
$ LD_LIBRARY_PATH=$(pwd) ./ctf WRONG_ARGUMENT
checking 'WRONG_ARGUMENT'
$ echo $?
1And we can now confirm that you need the specific arg show_me_the_flag in order to get the ok response in STDOUT.
-
Trace the libraries of the ctf binary
Let's use the
ltracecommand to trace the execution of the ctf binary:$ LD_LIBRARY_PATH=$(pwd) ltrace -i -C ./ctf show_me_the_flag [0x400fe9] __libc_start_main(0x400bc0, 2, 0x7ffda203f098, 0x4010c0 <unfinished ...> [0x400c44] __printf_chk(1, 0x401158, 0x7ffda203f4e7, 128checking 'show_me_the_flag' ) = 28 [0x400c51] strcmp("show_me_the_flag", "show_me_the_flag") = 0 [0x400cf0] puts("ok"ok ) = 3 [0x400d07] rc4_init(rc4_state_t*, unsigned char*, int)(0x7ffda203ee60, 0x4011c0, 66, 0x7fb67be48504) = 0 [0x400d14] std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >::assign(char const*)(0x7ffda203eda0, 0x40117b, 58, 3) = 0x7ffda203eda0 [0x400d29] rc4_decrypt(rc4_state_t*, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&)(0x7ffda203ee00, 0x7ffda203ee60, 0x7ffda203eda0, 0x7e889f91) = 0x7ffda203ee00 [0x400d36] std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >::_M_assign(std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > const&)(0x7ffda203eda0, 0x7ffda203ee00, 0x7ffda203ee10, 0) = 0x7ffda203edb0 [0x400d53] getenv("GUESSME") = nil [0xffffffffffffffff] +++ exited (status 1) +++
Notice the
getenv("GUESSME")towards the bottom of the output. This implies an environment variable namedGUESSMEcan be set and might affect the operation of the binary. Let's try to set it to something:$ LD_LIBRARY_PATH=$(pwd) GUESSME=1 ./ctf show_me_the_flag checking 'show_me_the_flag' ok guess again!
$ LD_LIBRARY_PATH=$(pwd) GUESSME=1 ltrace -i -C ./ctf show_me_the_flag ... [0x400d53] getenv("GUESSME") = "1" [0x400d6e] std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >::assign(char const*)(0x7ffd9e729a70, 0x401183, 5, 0xffffffe0) = 0x7ffd9e729a70 [0x400d88] rc4_decrypt(rc4_state_t*, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&)(0x7ffd9e729ad0, 0x7ffd9e729b10, 0x7ffd9e729a70, 0x401183) = 0x7ffd9e729ad0 [0x400d9a] std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >::_M_assign(std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > const&)(0x7ffd9e729a70, 0x7ffd9e729ad0, 0x23382f0, 0) = 0x23382a0 [0x400db4] operator delete(void*)(0x23382f0, 0x23382f0, 21, 0) = 0 [0x400dd7] puts("guess again!"guess again! ) = 13 [0x400c8d] operator delete(void*)(0x23382a0, 0x2337e70, 0x7f0f285a68c0, 0x7f0f284d3504) = 0 [0xffffffffffffffff] +++ exited (status 1) +++
While we can trace the execution of each function, we still can't get to the expected value.
-
Determine the expected value of GUESSME
$