Quick introduction to reverse engineering for beginners

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.text:005414A7 push ecx ; Filename .text:005414A8 call _fopen .text:005414AD mov ebp, eax .text:005414AF push ebp ; File .text:005414B0 push 1 ; Count .text:005414B2 push edi ; Size .text:005414B3 push esi ; Str .text:005414B4 call _fwrite .text:005414B9 push ebp ; File .text:005414BA call _fclose .text:005414BF push ebx ; Memory .text:005414C0 call _free .text:005414C5 add esp, 2Ch .text:005414C8 pop edi .text:005414C9 pop esi .text:005414CA pop ebp .text:005414CB pop ebx .text:005414CC retn .text:005414CC decrypt_file endp All function and label names are given by me while analysis. I started from top. Here is a function taking two file names and password. .text:00541320 ; int __cdecl crypt_file(int Str, char *Filename, int password) .text:00541320 crypt_file proc near .text:00541320 .text:00541320 Str = dword ptr 4 .text:00541320 Filename = dword ptr 8 .text:00541320 password = dword ptr 0Ch .text:00541320 Open file and report error in case of error: .text:00541320 mov eax, [esp+Str] .text:00541324 push ebp .text:00541325 push offset Mode ; "rb" .text:0054132A push eax ; Filename .text:0054132B call _fopen ; open file .text:00541330 mov ebp, eax .text:00541332 add esp, 8 .text:00541335 test ebp, ebp .text:00541337 jnz short loc_541348 .text:00541339 push offset Format ; "Cannot open input file!\n" .text:0054133E call _printf .text:00541343 add esp, 4 .text:00541346 pop ebp .text:00541347 retn .text:00541348 ; --------------------------------------------------------------------------- .text:00541348 .text:00541348 loc_541348: Get file size via fseek()/ftell(): .text:00541348 push ebx .text:00541349 push esi 161
.text:0054134A push edi .text:0054134B push 2 ; Origin .text:0054134D push 0 ; Offset .text:0054134F push ebp ; File ; move current file position to the end .text:00541350 call _fseek .text:00541355 push ebp ; File .text:00541356 call _ftell ; get current file position .text:0054135B push 0 ; Origin .text:0054135D push 0 ; Offset .text:0054135F push ebp ; File .text:00541360 mov [esp+2Ch+Str], eax ; move current file position to the start .text:00541364 call _fseek This piece of code calculates file size aligned to 64-byte border. This is because this cryptoalgorithm works with only 64-byte blocks. Its operation is pretty simple: divide file size by 64, forget about remainder and add 1, then multiple by 64. The following code removes remainder as if value was already divided by 64 and adds 64. It is almost the same. .text:00541369 mov esi, [esp+2Ch+Str] .text:0054136D and esi, 0FFFFFFC0h ; reset all lowest 6 bits .text:00541370 add esi, 40h ; align size to 64-byte border Allocate buffer with aligned size: .text:00541373 push esi ; Size .text:00541374 call _malloc Call memset(), e,g, clear allocated buffer 2 . .text:00541379 mov ecx, esi .text:0054137B mov ebx, eax ; allocated buffer pointer -> to EBX .text:0054137D mov edx, ecx .text:0054137F xor eax, eax .text:00541381 mov edi, ebx .text:00541383 push ebp ; File .text:00541384 shr ecx, 2 .text:00541387 rep stosd .text:00541389 mov ecx, edx .text:0054138B push 1 ; Count .text:0054138D and ecx, 3 .text:00541390 rep stosb ; memset (buffer, 0, aligned_size) Read file via standard C function fread(). .text:00541392 mov eax, [esp+38h+Str] .text:00541396 push eax ; ElementSize .text:00541397 push ebx ; DstBuf .text:00541398 call _fread ; read file .text:0054139D push ebp ; File .text:0054139E call _fclose Call crypt(). This function takes buffer, buffer size (aligned) and password string. 2 malloc() + memset() could be replaced by calloc() 162
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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$[esp+36] mov
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and r11, r8 and rsi, r8 and r10, r1
Page 116 and 117: pop rdi pop rsi ret 0 s1 ENDP Nothi
Page 118 and 119: 2.1 LEA instruction LEA (Load Effec
Page 120 and 121: 2.3 npad It’s an assembler macro
Page 122 and 123: 2.4 Signed number representations T
Page 124 and 125: call function function: .. do somet
Page 126 and 127: 3.2 String Debugging messages are o
Page 128 and 129: .text:3011E91B DD 1E fstp qword ptr
Page 130 and 131: loc_80483F2: pop ebp retn _f endp 4
Page 132 and 133: loc_8048444: loc_8048448: loc_80484
Page 134 and 135: public f2 f2 proc near push ebp mov
Page 136 and 137: loc_804840A: loc_804842E: loc_80484
Page 138 and 139: mov esi, DWORD PTR _k$[esp+16] push
Page 140 and 141: or eax, edx retn f endp 4.1.8 Task
Page 142 and 143: 4.2 Middle level 4.2.1 Task 2.1 Wel
Page 144 and 145: mov [edx+eax], bl mov ebx, edi mov
Page 146 and 147: loc_80485AB: ; CODE XREF: f+1E0 ; f
Page 148 and 149: mov [ecx+eax], dl inc eax cmp ebx,
Page 150 and 151: Chapter 5 Tools ∙ IDA as disassem
Page 152 and 153: Chapter 7 More examples 147
Page 154 and 155: .text:00541050 arg_0 = dword ptr 4
Page 156 and 157: .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 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 208 and 209: } 9.1.8 Task 1.8 Solution: two 100*
Page 210 and 211: #define PUSH(low, high) ((void) ((t
Page 212 and 213: } } ignore one or both. Otherwise,
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Quick introduction to reverse engineering for beginners

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