Quick introduction to reverse engineering for beginners

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} 9.1.8 Task 1.8 Solution: two 100*200 matrices of double type addition. C/C++ source code: #define M 100 #define N 200 void s(double *a, double *b, double *c) { for(int i=0;i
version 2.1 of the License, or (at your option) any later version. The GNU C Library is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU Lesser General Public License for more details. You should have received a copy of the GNU Lesser General Public License along with the GNU C Library; if not, write to the Free Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA. */ /* If you consider tuning this algorithm, you should consult first: Engineering a sort function; Jon Bentley and M. Douglas McIlroy; Software - Practice and Experience; Vol. 23 (11), 1249-1265, 1993. */ #include #include #include #include typedef int (*__compar_d_fn_t) (__const void *, __const void *, void *); /* Byte-wise swap two items of size SIZE. */ #define SWAP(a, b, size) \ do \ { \ register size_t __size = (size); \ register char *__a = (a), *__b = (b); \ do \ { \ char __tmp = *__a; \ *__a++ = *__b; \ *__b++ = __tmp; \ } while (--__size > 0); \ } while (0) /* Discontinue quicksort algorithm when partition gets below this size. This particular magic number was chosen to work best on a Sun 4/260. */ #define MAX_THRESH 4 /* Stack node declarations used to store unfulfilled partition obligations. */ typedef struct { char *lo; char *hi; } stack_node; /* The next 4 #defines implement a very fast in-line stack abstraction. */ /* The stack needs log (total_elements) entries (we could even subtract log(MAX_THRESH)). Since total_elements has type size_t, we get as upper bound for log (total_elements): bits per byte (CHAR_BIT) * sizeof(size_t). */ #define STACK_SIZE (CHAR_BIT * sizeof(size_t)) 204
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Quick introduction to reverse engin
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1.16 C++ classes . . . . . . . . .
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Preface Here (will be) some of my n
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1.1 Hello, world! Let’s start wit
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1.2 Stack Stack — is one of the m
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(_snprintf() function works just li
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1.3 printf() with several arguments
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1.4 scanf() Now let’s use scanf()
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GCC replaced first printf() call to
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}; return 0; printf ("What you ente
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1.5 Passing arguments via stack Now
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1.6 One more word about results ret
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push OFFSET $SG741 ; ’a
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1.8 switch()/case/default 1.8.1 Few
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1.8.2 A lot of cases If switch() st
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loc_804840C: ; DATA XREF: .rodata:0
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et 0 _main ENDP Nothing very specia
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add esp, 1Ch xor eax, eax ; return
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C/C++ standard defines char type as
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eax=eax+ecx return eax ... and this
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mov eax, ecx imul edx sar edx, 1 mo
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; current stack state: ST(0) = resu
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; in local stack here 8 bytes still
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fld QWORD PTR _a$[esp-4] ; current
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In other words, fnstsw ax / sahf in
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1.13 Arrays Array is just a set of
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mov [esp+70h+var_70], eax call _pri
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jge SHORT $LN1@main mov ecx, DWORD
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ebp 0x15 0x15 esi 0x0 0 edi 0x0 0 e
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1.14 Bit fields A lot of functions
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So, let’s download Linux Kernel 2
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There are some redundant code prese
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y 10 just by dropping last digit (3
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_key$ = 8 ; size = 4 _len$ = 12 ; s
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1.15 Structures It can be defined t
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call DWORD PTR __imp__GetSystemTime
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1.15.4 Fields packing in structure
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}; int b; struct outer_struct { cha
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}; printf ("stepping=%d\n", tmp->st
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call ___printf_chk mov eax, esi shr
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mov edx, DWORD PTR _t$[ebp] or edx,
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1.16 C++ classes I placed a C++ cla
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push ecx mov DWORD PTR _this$[ebp],
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call _ZN1c4dumpEv lea eax, [esp+20h
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1.17 Unions 1.17.1 Pseudo-random nu
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xor eax, eax pop esi mov esp, ebp p
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Let’s compile it in MSVC 2010 (I
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comp() function: public comp comp p
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Some compilers can do vectorization
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mov ecx, [esp+10h+var_10] mov edx,
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GCC In all other cases, non-SSE2 co
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add ebx, edx lea esi, [esi+0] loc_8
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movdqa xmm1, XMMWORD PTR [ecx+16] a
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1.20 x86-64 It’s a 64-bit extensi
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} x40 = a3 | x31; x41 = x24 & ~x37;
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or ebx, DWORD PTR _x6$[esp+36] mov
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and r11, r8 and rsi, r8 and r10, r1
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pop rdi pop rsi ret 0 s1 ENDP Nothi
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2.1 LEA instruction LEA (Load Effec
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2.3 npad It’s an assembler macro
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2.4 Signed number representations T
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call function function: .. do somet
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3.2 String Debugging messages are o
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.text:3011E91B DD 1E fstp qword ptr
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loc_80483F2: pop ebp retn _f endp 4
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loc_8048444: loc_8048448: loc_80484
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public f2 f2 proc near push ebp mov
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loc_804840A: loc_804842E: loc_80484
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mov esi, DWORD PTR _k$[esp+16] push
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or eax, edx retn f endp 4.1.8 Task
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4.2 Middle level 4.2.1 Task 2.1 Wel
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mov [edx+eax], bl mov ebx, edi mov
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loc_80485AB: ; CODE XREF: f+1E0 ; f
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mov [ecx+eax], dl inc eax cmp ebx,
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Chapter 5 Tools ∙ IDA as disassem
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Chapter 7 More examples 147
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.text:00541050 arg_0 = dword ptr 4
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.text:005410F3 .text:005410F3 loc_5
Page 158 and 159: .text:005411A9 pop ebx .text:005411
Page 160 and 161: .text:00541249 .text:00541249 next_
Page 162 and 163: .text:005412FC mov esi, offset cube
Page 164 and 165: .text:005413DC call _fwrite ; write
Page 166 and 167: .text:005414A7 push ecx ; Filename
Page 168 and 169: .text:005413A3 mov ecx, [esp+44h+pa
Page 170 and 171: .text:00541408 push offset aRb ; "r
Page 172 and 173: }; }; return; fseek (f, 0, SEEK_END
Page 174 and 175: .text:005411B5 mov ebp, eax Check f
Page 176 and 177: .text:0054124C inc ebp .text:005412
Page 178 and 179: This instruction clears bit, in oth
Page 180 and 181: .text:005410CD add esp, 40h .text:0
Page 182 and 183: }; for (i=0; i
Page 184 and 185: }; tmp[y][z]=get_bit (row, y, z); f
Page 186 and 187: }; fread (buf, flen, 1, f); fclose
Page 188 and 189: 7.2 SAP client network traffic comp
Page 190 and 191: .text:64413F68 .text:64413F68 loc_6
Page 192 and 193: .text:64405171 call dbg .text:64405
Page 194 and 195: .text:64404FF7 push offset aLogging
Page 196 and 197: .text:64405113 loc_64405113: .text:
Page 198 and 199: Now let’s dig deeper and find con
Page 200 and 201: .text:64411E0B .text:64411E0B loc_6
Page 202 and 203: Chapter 8 Other things 8.1 Compiler
Page 204 and 205: Chapter 9 Tasks solutions 9.1 Easy
Page 206 and 207: 9.1.5 Task 1.5 Hint #1: Keep in min
Page 210 and 211: #define PUSH(low, high) ((void) ((t
Page 212 and 213: } } ignore one or both. Otherwise,
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